PLANE   AND   SOLID 
GEOMETRX 


BY 

WEBSTER   WELLS,    S.B. 

AUTHOR   OF   A   8EHIES    OF   TEXTS   ON   MATHEMATICS 

AND 

WALTER   W.    HART,   A.B. 

ASSISTANT   PROFESSOR    OF   MATHEMATICS,    UNIVERSITY    OF   WMO^WIN 
COURSE    FOR    THE    TRAINING    OF    TEACHERS 


o>Hc 


D.    C.    HEATH    &   COMPANY,   PUBLISHERS 
BOSTON  NEW  YORK  CHICAGO 


aA4   3 


WELLS   AND   HAKT'S   MATHEMATICS 


First  Year  Algebra 

A  one  year  course 

New  High  School  Algebra 

A  three  semester  course 

Second  Course  in  Algebra 

Appears  in  a  brief  and  in  an  enlarged  edition. 
Suitable  for  a  one  or  a  two  semester  course  following 
a  first  year  course 

Plane  Geometry 

Provision  is  made  for  a  brief  or  an  extended  course 

Solid  Geometry 


D.  C.  HEATH  &  CO.,  PUBLISHERS 

EDUcAT«ojv<  i)epf: 

Copyright,  1915,  1916, 
By  D.  C.  Heath  &  Co. 

I  a6 


PREFACE 

Greek  Geometry,  the  finest  product  of  deductive  thinking 
which  high  school  pupils  encounter,  has  come  down  to  us 
through  twenty-two  centuries  practically  unchanged  in  essen- 
tial content  or  form.  It  has  been  presented  in  texts,  each 
built  upon  a  preceding  one  and  each  good  in  its  day,  which 
have  sought  to  present  the  great  science  in  accord  with  the 
ideals  of  their  time. 

This  text  is  a  thorough  revision  of  Wells's  Essentials  of 
Geometry  in  accord  with  current  scientific  and  pedagogical 
thought.  The  scientific  ideal  is  represented  the  world  over 
by  Hilbert's  Foundations  of  Geometry.  (Translated  by  Town- 
send,  Open  Court  Pub.  Co.,  Chicago.)  The  pedagogical  ideals 
are  represented  in  this  country  by  the  Report  of  the  National 
Committee  of  Fifteen.  (See  Mathematics  Teacher,  Dec,  1912  ; 
School  Science  and  Mathematics,  1911 ;  Proceedings  of  N.  E.  A., 
1911.)  These  ideals  and  the  personal  experience  of  one  of  the 
authors  in  teaching  high  school  geometry  in  recent  years  have 
been  the  determining  factors  in  the  making  of  this  text. 
Permit  us  to  direct  attention  to  some  of  its  features. 

In  each  Book,  the  fundamentally  important  theorems  are 
given  first.  These  theorems  present  a  safe  and  sane  minimum 
course.  These  are  followed  in  each  Book  by  one  or  more 
groups  of  theorems  or  applications  which  are  strictly  supple- 
mentary,—  material  which  either  has  long  appeared  in  geom- 
etries in  some  form  or  has  been  introduced  in  recent  years  to 
add  to  the  pupils'  interest.     Teachers  will  find  no  difficulty  in 

54  X  an  8 


IV  PREFACE 

making  selection  from  this  material,  and,  on  the  other  hand, 
will  not  be  embarrassed  by  omitting  any  of  it.  (See  pp.  172, 
210,  and  245.) 

The  introduction  presents  only  the  immediately  necessary 
concepts,  notation,  and  terminology.  Emphasis  is  upon  the 
acquisition  of  these  and  of  skill  in  the  use  of  tools,  and  above 
all  upon  the  acquisition  of  the  important  point  of  view  pre- 
sented in  §§  48-50. 

The  fundamental  constructions  are  placed  early  in  Book  I 
so  that  pupils  can  be  required  to  construct  their  figures  ;  they 
are  not  placed  earlier  because  they  cannot  be  proved  earlier. 

Authorities  and  details  of  demonstrations  which  pupils  can 
supply  are  increasingly  omitted  from  the  demonstrations,  and 
often  only  suggestions  are  given.  The  resulting  proofs  are  an 
incentive  to  real  thinking  for  all  the  members  of  the  class ; 
they  do  not  consume  time  that  can  be  spent  more  profitably 
upon  exercises  and  other  valuable  supplementary  material. 

Pupils  are  encouraged  to  plan  their  proofs  instead  of  plung- 
ing blindly  into  a  demonstration.     (See  §§69  and  117.) 

Unnecessary  corollaries  have  been  omitted,  and  dignity  and 
importance  is  given  to  those  which  are  included  in  the  text. 
(See  §§  71,  96,  101.) 

The  stages  of  the  proof  are  plainly  marked,  the  steps  are 
numbered,  the  reasons  are  given  in  full,  and  the  proofs  are 
arranged  attractively  on  the  page. 

Carefully  selected  exercises  follow  most  of  the  propositions. 
Notice  exercises  such  as  Exercises  2,  23,  45,  63,  of  Book  I,  de- 
signed to  teach  concretely  and  inductively  the  theorems  which 
immediately  follow.  Notice  also  the  illustrative  exercises 
which  set  a  standard  for  the  pupils'  solution.  (See  pp.  31,  32, 
157.)  Enough  exercises  are  provided  for  a  minimum  course. 
Besides  these,  there  are  miscellaneous  exercises  at  the  close  of 
each  Book,  depending  upon  only  the  theorems  of  the  minimum 
course.  Finally  there  will  be  found  from  time  to  time  a  note 
like  that  on  page  52,  referring  to  supplementary  exercises  at 
the  end  of  the  text.     (See  pp.  52,  59,  83.)     Suggestions  are 


PREFACE  V 

given  with  exercises  where  experience  has  shown  that  a  ma- 
jority of  a  class  require  such  assistance  in  order  to  do  effective 
work.     (See  Book  1,  Ex.  128,  131,  etc.) 

Simple  applied  problems  (see  Book  I,  Ex.  15,  37,  39,  40,  41, 
etc.)  and  artistic  designs  (see  pp.  1,  47,  50,  etc.)  exhibit  to  the 
pupils  some  of  the  uses  of  geometry.  Only  simple  applica- 
tions are  included  in  the  minimum  course.  Other  applications 
are  introduced  among  the  supplementary  exercises  at  the  end 
of  the  text  and  among  the  supplementary  topics  at  the  close 
of  certain  of  the  Books.     (See  pp.  138,  172,  174,  246,  etc.) 

A  brief  history  of  geometry  is  included  in  the  introduction, 
and  other  historical  references  are  introduced  from  time  to  time 
throughout  the  text.     (See  pp.  29,  36,  46,  240,  etc.) 

Axioms  are  defined  in  the  accepted  modern  form  (p.  22). 
They  are  introduced  only  as  they  become  necessary.  (See 
pp.  22,  29,  50,  82.)  In  the  introduction,  their  meaning  is  made 
clear  by  suitable  preliminary  exercises.  (See  Introduction,  Ex. 
22,  23,  24,  37,  etc.)  The  definitions  also  are  modern  and  con- 
sistent,'even  though  they  are  in  some  cases  different  from  those 
ordinarily  given.  (See  §  §  1,2,  4,  5,  47,  the  note  on  p.  27,  etc.) 
For  example,  after  defining  a  circle  as  a  line,  which  is  correct, 
there  is  every  reason  for  also  defining  a  polygon  as  a  line, 
instead  of  defining  it  as  a  portion  of  a  plane.  It  may  seem 
strange  at  first  also  not  to  find  in  the  first  paragraph  of  the 
text  the  attempted  distinction  between  a  physical  and  a  geo- 
metrical solid,  —  something  that  is  psychologically  impossible 
for  beginners,  —  but  the  authors  believe  firmly  that  there  is 
much  to  recommend  their  own  informal  statements  in  §  1. 

The  incommensurable  cases  are  dismissed  with  a  mere  re- 
mark on  pages  113,  149,  and  194,  and  are  treated  fully  only 
after  the  theory  of  limits  is  given  on  page  260. 

The  mensuration  of  the  circle  is  treated  informally  at  first 
on  page  238.  The  treatment  involves  nevertheless  the  basic 
ideas  which  are  developed  more  fully  in  the  formal  treatment 
of  the  same  topic  which  appears  as  one  of  the  supplementary 
topics  of  Book  V  on  page  248.     This  treatment  is  as  elemen- 


VI  PREFACE 

tary  as  the  difficulty  of  the  subject  permits  ;  to  give  less  would 
render  the  treatment  either  incomprehensible  or  incomplete. 
On  the  other  hand,  the  treatment  is  as  sound  as  an  elementary 
presentation  renders  possible ;  to  give  more  would  certainly 
render  the  subject  distasteful  to  an  average  high  school  class. 

In  the  treatment  of  the  mensuration  of  the  cylinder  and 
the  cone,  the  fundamental  limits  theorems  are  assumed  on 
the  ground  that  rigorous  proofs  are  beyond  the  scope  of  an 
elementary  course.  In  the  enunciation  of  the  area  theorems 
for  portions  of  the  surface  of  a  sphere,  changes  have  been 
made  which  enable  pupils  both  to  learn  and  to  remember  the 
theorems  more  readily. 

The  course  in  Solid  Geometry  is  practical  in  the  sense  that 
the  mensuration  theorems  for  the  common  solids  are  given  the 
place  of  prominence.  Por  example,  in  Book  IX  the  mensu- 
ration of  the  sphere  is  treated  in  the  minimum  course,  —  the 
mathematically  interesting  theorems  about  spherical  geometry 
being  grouped  as  a  supplementary  topic.  Besides  this  empha- 
sis given  to  the  mensuration  theorems,  some  natural  applica- 
tions of  solid  geometry  are  touched  upon  in  the  exercises. 


CONTENTS 

PAOB 

Introduction 1 

Book  I.     Rectilinear  Figures 29 

Supplementary  Theorems 87 

Miscellaneous  Exercises 91 

Book  II.    The  Circle 93 

Measurement  of  Angles  and  Arcs 1 12 

Loci 124 

Supplementary  Topics 128 

Miscellaneous  Exercises 139 

Book  III.     Proportion  —  Similar  Polygons 141 

Supplementary  Topics 172 

Miscellaneous  Exercises 189 

Book  IV,     Areas  of  Polygons 191 

Supplementary  Topics 210 

Miscellaneous  Exercises 219 

Book  V.     Regular  Polygons 221 

Mensuration  of  the  Circle  —  Informal  Treatment        .        .        .  238 

Supplementary  Topics 245 

Supplementary  Exercises  —  Book  I 273 

Supplementary  Exercises  —  Book  II 283 

Supplementary  Exercises  —  Book  III 289 

Supplementary  Exercises  —  Book  IV 294 

Supplementary  Exercises  —  Book  V 299 


vn 


viii  CONTENTS 

PAGE 

Book  VI.     Lines  and  Planes  —  Polyedral  Angles    ....  307 

Supplementary  Topics o  339 

Supplementary  Theorems    .        .        .        .        o        .        .        .  343 

Book  VII.     Polyedra        ........        e  346 

Supplementary  Topics .        .  376 

Miscellaneous  Theorems      ....        ....  380 

Book  VIII.     The  Cylinder  and  the  Cone .        .        .        .        ,        .388 

Measuring  the  Cylinder 392 

The  Cone 396 

Supplementary  Topics 405 

Book  IX.     The  Sphere 409 

Spherical  Polygons 415 

Volume  of  a  Sphere .  434 

Supplementary  Topics 438 

Supplementary  Exercises  —  Books  VI  to  IX 454 

Index 463 


PLANE   AND   SOLID   GEOMETRY 


SYMBOLS 


= ,  is  equal  to ;  equals. 
>,  is  greater  than. 
<,  is  less  than. 
II  ,  is  parallel  to ;  parallel. 
±,  is  perpendicular  to. 
J_,  perpendicular. 
~,  is  similar  to. 
^,  is  congruent  to. 

Any   symbol    representing 
plural  by  affixing  the  letter  s 


Z,  angle. 

A,  triangle. 

O,  parallelogram. 

□,  rectangle. 

O,  circle. 

.'.,  therefore. 

=  ,  is  identically  equal  to. 

=  ,  approaches  as  limit. 

a   noun   is    converted  into 
thus  A  means  angles. 


the 


ABBREVIATIONS 


Adj., 

adjacent. 

Ex., 

exercise. 

Alt, 

alternate. 

Ext., 

exterior. 

Ax., 

axiom. 

Hom.^ 

homologous. 

Con., 

conclusion. 

Hyp/, 

hypothesis. 

Cong., 

congruent. 

Int., 

interior. 

Const., 

construction. 

Rect., 

rectangle. 

Cor., 

corollary. 

Rt, 

right. 

Corres. 

,  corresponding. 

St., 

straight. 

Def., 

definition. 

Supp. 

,  supplementary. 

It  is  not  necessary  to  learn  any  of  these  symbols  until  they 
are  introduced  in  the  text. 


m-  i( 


m 


iWg^' 


ifH 


A  View  in  the  Congressional  Library,  Illustrating  the  Use 
OF  Geometry  in  Architecture 


GEOMETRY 


INTRODUCTION 

In  arithmetic  and  algebra,  frequent  reference  is  made  to 
the  rectangle,  the  square,  the  triangle,  and  the  circle.  These 
are  geometrical  figures,  and  in  geometry  a  careful  study  of 
them  and  of  many  others  is  made. 

Geometrical  figures  are  used  constantly  in  architecture. 


Plans  of  a  House 
They  often  form  the  basis  of  artistic  designs. 


A  Textile  Pattern 


An  Autistic  Tbay 


2  PLANE   GEOMETRY 

Our  playgrounds  are  often  laid  out  in  geometrical  forms. 


A  Football  Field 

Familiarity  with  such  figures  and  their  properties,  and 
ability  to  construct  and  measure  them,  is  both  interesting  and 
worth  while.  It  is  interesting  also  to  know  how  man  has 
developed  his  knowledge  of  such  figures  and  his  skill  in  using 
them. 

HISTORY  OF  GEOMETRY 

Geometry  as  it  is  now  studied  has  been  handed  down  to  us 
from  the  Greeks.  The  word  "  geometry  "  is  derived  from  two 
Greek  words  meaning  the  earth  and  to  measure;  this  fact  is 
evidence  that  the  Greeks  believed  that  geometry  was  inti- 
mately associated  with  or  else  had  been  developed  out  of  the 
practical  business  of  measuring  the  earth,  —  surveying. 

The. Greeks  receiyed  thei?  start  in  geometry  from  the  Egyp- 
tians, 'tkales  Qf'*3^iiilcs''(.feO-550  b.c.)  is  given  special  credit 
for  tBans<plaiut4iig»arknowledgdi3f  Egyptian  geometry  to  Greece. 

BitJ-  IKe  Egyptians*  ©r?giimte'*geometry  ?  Whether  they  did 
or  not,  there  is  evidence  that  they  had  some  knowledge  of  prac- 
tical geometry.  Their  pyramids  and  other  marvelous  struc- 
tures point  to  this  fact.  Also,  there  is  in  the  British  Museum 
a  papyrus  written  about  1700  b.c.  by  an  Egyptian,  commonly 


INTRODUCTION  3 

called  Ahmes,  which  contains  among  other  interesting  mathe- 
matical records  some  formulae  for  measuring  geometrical 
figures.  This  papyrus  is  a  copy  of  another  written  before  the 
time  of  Ahmes.  Herodotus,  a  Greek  traveler  and  historian, 
is  said  to  be  responsible  for  the  story  that  the  Egyptians  de- 
veloped these  rules  of  mensuration  because  of  the  necessity  of 
frequently  surveying  the  lands  which  were  inundated  by  the 
floods  of  the  Nile.  The  Egyptians  must  have  obtained  their 
formulae  by  experiment  or  by  observation.  Some  of  the 
formulae  were  incorrect  and  their  formula  for  measuring  the 
area  of  a  circle  was  less  accurate  than  that  developed  later  by 
the  Greeks. 

The  Greeks  became  interested  in  geometry  for  its  own  sake 
as  well  as  for  its  usefulness.  In  the  three  hundred  years 
following  the  time  of  Thales,  geometry  grew  into  a  great 
science  in  their  schools,  far  exceeding  the  geometry  of  the 
Egyptians  in  the  number  and  interest  of  the  facts  discovered, 
and  in  the  accuracy  and  usefulness  of  the  results.  Pythag- 
oras and  Plato  were  the  leaders  of  two  groups  of  students 
which  were  responsible  for  much  of  the  advance  made  in  the 
subject. 

Hippocrates  (about  420  b.c.)  made  an  attempt  to  prepare  a 
text  on  geometry,  but  it  remained  for  Euclid  to  write  what  be- 
came the  standard  text.  Euclid  lived  between  330  and  275  b.c. 
He  was  one  of  the  first  and  greatest  mathematicians  who 
taught  at  the  University  of  Alexandria.  As  a  teacher  he  felt 
the  need  of  a  text  by  which  to  lead  beginners  through  the 
known  facts  of  elementary  geometry.  He  therefore  gathered 
together  and  systematized  these  facts  in  a  book  known  as  the 
Elements.  Euclid's  Elements  has  stood  as  the  model  for  all 
subsequent  texts  on  the  subject. 

During  the  two  thousand  years  since  the  time  of  Euclid, 
geometry  has  been  studied  by  all  civilized  peoples  and  has  been 
enriched  from  time  to  time  by  their  mathematicians.  This 
history  is  so  long  and  the  details  are  so  technical  that  it  is 
unwise  to  attempt  to  give  more  of  it  at  this  time. 


PLANE   GEOMETRY 


4y 


INFORMAL  PREPARATORY  GEOMETRY 

1.  The   adjoining  figure   is   a  cube.     It  has    six   surfaces. 
Each    surface   is   bounded    by   four    lines, — 
straight  lines.     Each  straight  line  is  bounded 
by  two  points. 

The  surfaces  of  a  cube,  which  are  smooth 
and  flat,  are  called  Plane  Surfaces;  they  are 
such   that   a   straightedge   (ruler)   will    touch 
the  surface  at  all  points  of  the  straightedge,  no  matter  where 
the  plane  surface  may  be  tested. 

2.  Plane  Geometry  is  the  study  of  figures  like  the  square, 
the  triangle,  the  circle,  etc.,  — figures  which  lie  in  a  plane 
surface. 

A  Plane  Geometrical  Figure  is  a  combination  of  points  and 
lines  which  lie  in  one  plane  surface.  Only  such  figures  are 
considered  in  plane  geometry. 

Ex.  1.  Test  the  surface  of  your  desk  with  your  ruler  to  determine 
whether  the  surface  is  a  plane  or  not.     (See  §  1.) 

Ex.  2.     What  are  some  other  objects  which  have  plane  surfaces  ? 

Ex.  3.  How  do  men  who  are  laying  a  concrete  walk  make  use  of  this 
test  in  order  to  make  the  surface  of  the  walk  approximately  plane  ? 


3.   Solid  Geometry  is  the  study  of  figures  like  the  cube,  the 
sphere,  the  cylinder,  the  pyramid,  the  cone,  etc. 


Pyramid 


Cylinder 


Cone 


INTRODUCTION  5 

4.  A  Point  is  represented  to  the  eye  by  a  small  dot.  .  A 

A  point  is  named  by  placing  beside  it  a  capital  printed  letter;  as 
point  A. 

A  point  represents  position  only. 

5.  A  Straight  Line  is  represented  to  the  eye  by  a  mark  made 
by  drawing  a  pencil,  a  pen,  or  a  piece  of  crayon  along  the  edge 
of  a  straightedge, 

A  line  represents  length  only. 

A  Curved  Line  is  a  line  no  part  of  which  is  straight. 

A  Broken  Line  is  a  line  composed  of  different  successive 

straight  lines. 

F        Q 


Straight  Line  Curved  Line  Broken  Line 

6.  Lines  like  the  adjoining  ones  are  called  closed  lines. 

It  is  apparent  that  a  closed  line  incloses 
a  portion  of  the  plane. 

7.  The  word  "  line  "  will  mean  a 
straight  line  hereafter  unless  otherwise  specified. 

Ex.  4.  Place  upon  paper  a  single  point,  (a)  Draw  through  it  one 
straight  line.  (6)  Can  you  draw  through  it  another  straight  line  ?  (c)  A 
third  ?     {(l)  How  many  straight  lines  can  be  drawn  through  one  point  ? 

Ex.  5.  Place  upon  paper  a  point  A  and  a  point  B.  (a)  Draw  from 
.4  to  ^  a  straight  line.  (6)  What  happens  when  you  try  to  draw  a  sec- 
ond straight  line  from  A  to  B?  (c)  How  many  different  straight  lines 
do  you  conclude  can  be  drawn  between  two  points  ? 

Ex.  6.  Can  more  than  one  curved  line  be  drawn  between  two  points  ? 
Illustrate. 

Ex.  7.  (a)  When  walking  along  a  straight  line,  are  you  moving  con- 
stantly in  the  same  direction  or  not  ?  (6)  Answer  the  same  question  if 
you  are  walking  along  a  curved  line. 

Ex.  8.  Draw  a  straight  line  2  inches  long.  Extend  it  one  inch  in 
each  direction. 


6  PLANE   GEOMETRY 

8.  It  will  be  assumed  as  apparent  from  the  preceding  exer- 
cises that : 

(a)   One  and  only  one  straight  line  can  be  drawn  through 

two  points. 
This   fact  is  also  expressed   thus :   two  points  determine  a 

straight  line, 
(h)  A  straight   line   can    he  extended   indefinitely   in  each 

directioyi. 

9.  The  straight  line  determined  by  points  A  and  B  is  called 

the  line  AB. 

A  B 


Ex.  9.  Select  three  points  which  are  not  in  one  straight  line.  Letter 
them  A,  J5,  and  C.  Draw  the  different  straight  lines  determined  by  them 
taken  two  at  a  time.     Name  the  straight  lines  that  you  get. 

Ex.  10.  If  four  towns  are  situated  so  that  no  three  can  be  connected 
by  one  straight  road,  how  many  roads  must  be  constructed  if  each  town 
is  to  be  connected  with  each  of  the  others  by  a  straight  road  ?  Illustrate 
by  a  drawing. 

Ex.  11.  Draw  the  straight  line  determined  by  two  points.  Then  turn 
the  straightedge  over,  and  again  draw  a  straight  line  between  the  two 
points.  If  the  edge  is  a  true  straightedge,  the  two  straight  lines  will 
coincide  (form  one  line).     Why  is  this  so  ? 

Ex.  12.  Make  a  straightedge  by  folding  a  piece  of  paper.  Test  it  by 
the  method  suggested  in  the  preceding  exercise. 

Ex.  13.  In  order  to  walk  across  a  field  in  a  straight  line,  a  boy  selects 
two  objects  which  are  in  the  direction  in  which  he  wishes  to  go,  one  of 
them  directly  between  him  and  the  other.  As  he  walks,  he  constantly 
keeps  the  first  object  between  himself  and  the  second. 

(a)  Why  can  he  guide  himself  in  this  manner  ? 

(6)  What  two  points  determine  the  straight  line  along  which  he  walks  ? 

10.   Two   lines,  straight  or   curved,     ^         ^^^ ^^/ 

intersect   if   they  have   one    or    more 

common  points.  The  common  points  are  called   Points   of 

Intersection. 


INTRODUCTION 

11.    Two  straight  lines  can  intersect  at  only  one  point 

If  they  were  to  intersect  in  two  points,     ^^  ^7) 

there  would  be  two  straight   lines  through 


these    two    points,  and   this  is   impossible        ^^^^ 
(§8).  <^^  ^^ 

This  fact  is  also  stated  thus :  two  intersecting  straight  lines 
determine  a  point. 

Ex.  14.  Draw  three  straight  lines  intersecting  by  pairs  which  do  not 
all  pass  through  one  point.     How  many  points  do  they  determine  ? 

Ex.  15.  If  there  are  in  a  county  four  straight  roads,  each  of  which 
crosses  each  of  the  others,  and  no  three  of  which  meet  at  one  point,  how 
many  crossings  are  there  ?    Illustrate.    C 

Ex.  16.  How  many  points  are  determined  by  five  straight  lines  in- 
tersecting by  pairs,  no  three  of  which  pass  through  a  common  point  ?     /  O 

Ex.  17.  Can  you  make  any  definite  statement  about  the  number  of 
points  of  intersection  of  two  curved  lines  ? 

12.  A  Line-segment  or  Segment  is  the  part  of 

a  straight  line  between  two  points  of  the  line;   ^ ^ 

as,  segment  RS. 

13.  Two  segments  are  equal  if  they  can  be  placed  so  that 
the  ends  of  the  one  are  exactly  upon  the  ends  of  the  other. 

The  tool  for  testing  the  equality  of  two  segments  is  the  dividers, 
rhe  dividers  are  spread  until  the  points  are  upon  A 

and  B  respectively.     If  the  dividers  can  then  be  placed    -^ E. 

with  their  points  on  C  and  D  respectively  without  chang- 
ing the  position  of  the  legs  of  the  dividers,  then  the  two 

segments  are  equal. 

AB  is  less  than  (<)  CD  if  AB  equals  a  part  of  CD. 

Ex.  18.  Determine  by  means  of  the  dividers  the 
relative  lengths  of  AB  and  BC;  of  ^5  and  CD;  of 
AB  and  AD. 

Ex.  19.     Draw  any  segment  AB.     On  a  line  of 
indefinite  length,  mark  off  from  a  point  0  of  that  line  a  segment  equal  to 
2  AB  ;  also  one  equal  to  3  AB. 


8  PLANE   GEOMETRY 

Ex.  20.     Draw  segments  AB  and  CD,  with  AB  greater  than  CD. 
(a)  On   a  line   of    indefinite    length,    mark    off    a  segment  equal  to 
AB  +  CD.     (6)  Mark  off  a  segment  equal  to  AB  -  CD. 

Ex.  21.    Let  AB  and  CD  be  two  segments.     Sup-  ^  ^ 

pose  that  AB  is  placed  upon  CD  with  point  A  on  point  ] — 

C.     (a)  Where  will  B  fall  if  AB=  CD?  C ^ 

(6)  Where,  if  ^^  =  i  CD? 

(c)  Where,  if  AB  is-  greater  than  CD  ? 

Ex.  22.  Suppose  that  two  segments  are  each  equal  to  a  third  seg- 
ment.    How  do  these  two  segments  compare  with  each  other  ? 

Ex.  23.  Suppose  that  two  segments  are  each  equal  to  equal  seg- 
ments.    How  do  these  segments  compare  with  each  other  ? 

Ex.  24.     Complete  the  following  sentences : 

(a)  If  equal  segments  are  added  to  equal  segments,  the  sums  are  ••• 
(6)  If  equal  segments  are  subtracted  from  equal  segments,  the  re- 
mainders are  •  •  • 

14.  It  will  be  assumed  as  apparent  that: 
the  straight  line-segment  is  the  shortest  line 
between  two  points.  \n 

The  Distance  between  two  points  is  the  length  of  the  seg- 
ment of  the  straight  line  between  the  points. 

To  obtain  a  straight  line  between  two  points,  a  carpenter  stretches  a 
piece  of  twine  between  the  two  points.  In  doing  so,  he  assumes  that  the 
shortest  line  between  two  points  is  the  straight  line. 

Ex.  25.     Why  are  streets  usually  made  straight  ? 

Ex.  26.     Why  do  people  often  "  cut  across  "  a  vacant  corner  lot  ? 

Ex.  27.  Place  upon  paper  points  A,  B,  and  C  so  that  they  do  not  all 
lie  upon  a  straight  line.  Draw  segments  AB,  BC,  and  AC.  By  means 
of  your  dividers  compare  the  longest  segment  with  the  sum  of  the  other 
two  segments. 

15.  A  point  bisects  a  segment  if  it  divides  the  segment  into 
two  equal  segments.  The  point  is  called  the  Mid-point  of  the 
segment. 

A  n  G  S  5 

Thus,  C  bisects  AB  ii  AC  =  CB. 


INTRODUCTION  9 

It  will  be  assumed  as  apparent  that  a  segment  has  only  one 
mid-point. 

Ex.  28.  Determine  with  your  dividers  whether  C  does  actually  bisect 
AB.  If  it  does,  what  part  of  AB  is  AC?  Does  D  bisect  AC?  Does 
E  bisect  CB  ?     (See  Fig.  §  15.) 

Ex.  29.  Draw  a  segment  of  any  length  and  locate  freehand  the 
point  which  you  think  bisects  the  segment.  Test  the  two  parts  to 
determine  whether  you  actually  located  the  mid-point  of  the  segment. 
(Continue  this  exercise  until  you  can  approximately  bisect  a  segment  in 
this  manner.) 

Ex.  30.     What  must  be  true  about  halves  of  equal  segments  ? 

16.  A  Circle  is  a  closed  curved  line  all  points  of    which 
are  equidistant  from  a  point  within 
called  the  Center. 

A  Radius  of  a  circle  is  the  distance 
from  the  center  to  any  point  on  the 
circle;  as  OA. 

A  Diameter  of  a  circle  is  a  seg- 
ment drawn  through  the  center  of. 
the  circle  with  its  ends  on  the  circle ; 
as  BD. 

A  Chord  of  a  circle  is  the  segment  joining  any  two  points 
of  the  circle  ;  as,  chord  CE. 

A  circle  can  be  drawn  with  any  point  as  center  and  any 
given  segment  as  radius. 

17.  Two  circles  having  equal  radii  can  be  made  to  coincide 
and  are  called  equal  circles.     Hence : 

All  radii  of  the  same  circle  or  of  equal  circles  are  equal. 
Ex.  31.     Draw  a  circle  of  radius  1  inch. 

Ex.  32.  Draw  two  circles  having  the  same  center  with  radii  of  1.5  in. 
and  2  in.  respectively. 

Ex.  33.  Draw  a  circle  and  a  straight  line  which  intersects  it.  How 
many  points  of  intersection  are  there  ? 

Ex.  34.  Draw  two  circles  that  intersect.  How  many  points  of 
intersection  are  there  ? 


10  PLANE   GEOMETRY 

18.   Circles  form  the  basis  of  numerous  designs. 


Can  you  copy  any  of  these  designs  ? 

19.  A  Half-line  or  Ray  is  the  part  of  a  straight  line  in  one 

direction  from  a  given  point  on  the  line;  as   ^ _^ 

p         in 
ray  m.  ^ 

Ex.  35.  How  many  end-points  has  a  line-segment  ?  A  ray  ?  A 
line? 

20.  An  Angle  (Z)  is  the  figure  formed  by  two  rays  drawn 
from  the  same  point.  ^B 

This  definition  was  introduced  by  a  mathema- 
tician, Beitrand,  in  1778.  q^ 1 ^j^ 

The  common  point  is  called  the  Vertex  of  the  angle. 

The  two  rays  are  called  the  Sides  of  the  angle. 

One  may  imagine  a  ray  starting  from  the  position  OA  and 
turning  about  point  O  until  it  occupies  the  position  OB.  OA 
is  then  called  the  initial  line,  and  OB  the  termiyial  line. 

Note  1.  —  The  size  of  an  angle  may  be  thought  of  as  depending  upon 
the  amount  of  turning  about  the  vertex  which  a  line  must  do  to  pass  from 
the  initial  line  to  the  terminal  line. 

Note  2.  —  The  portion  of  the  plane  over  which  the  line  would  pass 
is  said  to  be  within  the  angle.     (See  Note  2,  page  28.) 


INTRODUCTION  11 

Note  3.  —  Since  the  rays  extend  indefinitely,  it  is  clear  that  the  size 
of  an  angle  does  not  depend  upon  the  length  of  its  sides. 

An  angle  may  be  named  by  the  letter  at  its  vertex  if  there 
is  in  the  figure  only  one  angle  having  that  vertex  ;  as  Z  0. 

An  angle  may  be  indicated  by  a  number  placed  within  the 
angle  near  its  vertex  ;  as  Z  1. 

An  angle  may  be  named  by  reading  the  letters  A^  0,  B-,  as 
Z  AOB,  where  the  vertex  letter  0  is  placed  between  the  other 
two  letters. 

21.  Two  angles  are  equal  if  they 
can  be  made  to  coincide. 

Thus,  ii  ZE  can  be  placed  upon  Z  B 
so  that  point  E  is  on  point  J5,  line  ED 
on  line  BA,  and  line  EF  on  line  BC\ 
then  Z  E  equals  Z  B. 

22.  An  Z  AOB  is  less  than  Z  AOC  if  it  /^w^^ 
equals  a  part  of  Z  AOC.  0^^^^^ A 

Ex.  36.  Make  on  thin  paper  a  tracing  of  Z  DEF  (§  21).  Place  the 
tracing  over  Z  ABC  and  thus  determine  whether  the  angles  are  actually 
equal. 

Ex.  37.  "What  must  be  true  about  two  angles  each 
of  which  is  equal  to  a  third  angle  ? 

Ex.  38.  In  the  adjoining  figure,  make  a  tracing  of 
angle  1.  Determine  which  of  the  other  angles  are  equal 
to  Z  1. 

23.  A  line  bisects  an  angle  if  it  divides  the  angle  into  two 
equal  angles. 

Thus,  00  bisects  Z  AOB  ii  Z\  =  Z2. 

The  line  is  called  the  Bisector  of  the  angle. 
It  will   be  assumed   as  apparent   that  an    q^^ 
angle  has  only  one  bisector. 

Ex.  39.  Make  a  tracing  of  i^  1  and  determine  whether  Zl  is  actually 
equal  to  Z  2.     Is  00  actually  the  bisector  oiZAOB? 


12 


PLANE  GEOMETRY 


Ex.  40.     What  part  oiZAOBiaZl?     (See  Fig.  §  23.) 

Ex.  41.     Draw  any  angle.     Can  you  fold  the  paper  so  that  the  crease 
will  bisect  the  angle  ? 

Ex.  42.     Draw   any  angle.  Draw  a  line   which  you  think  bisects 

the  angle.    Test  the  equality  of  the  two  parts  of  the  angle  by  means  of 

tracing  paper.     (Continue  this  exercise  until  you  can  approximately 
bisect  an  angle  in  this  manner.) 

Ex.  43.     What  must  be  true  about  halves  of  equal  angles  ? 

Ex.  44.    Suppose  that  Z  ABC  is  placed  upon  Z  DEF  so  that  point  B 
is  on  point  E,  and  line  BA  is  on  line  ED. 

Where  will  BC  fall  if  ZABC  is  equal  to  /C  /F 

ZDEF'f 

Where  will  BC  fall  if  Z^^C  is  less  than 
ZDEF7 

Where  will  BC  fall  if  ZABC  is  greater  than  ZDEF7 

Ex.  45.     Complete  the  following  sentences  : 

(a)  If  equal  A  are  added  to  equal  zi,  the  sums  are  ••• 

(6)  If  equal  A  are  subtracted  from  equal  A^  the  remainders  are 

24.  Adjacent  Angles  are  two  angles  that  -^ 

liave  a  common  vertex  and  a  common  side 
between  tliem. 

Thus,  Z  1  and  Z  2  are  adjacent  angles. 

Ex.  46.     In  the  adjoining  figure : 
(a)  Ts  Z  1  adjacent  to  Z  2  ?     Why  ? 
(&)   Is  Z  2  adjacent  to  Z  3 


Why 


25.   Two  adjacent  angles  are  readily  added,    j^ 

Thus,  Z  AOB  +  Z  BOC  =  Z  AOC. 
Also,    ZAOC-ZBOC=ZAOB. 

Ex.  47.     In  the  figure  (§  25),  read  the  angle  which  represents : 
(a)  ZAOB  +  ZBOD;  (d)  ZBOE-ZDOE; 


(6)  ZBOC  +  ZCOD; 

(c)  ZBOC+ZCOD  +  ZDOE] 


(e)  ZAOC  +  ZCOD-ZBOD; 
(/)  ZAOE-ZDOE- ZCOD. 


INTRODUCTION 


13 


26.   If   one   straight   line   meets    another   straight   line   so 
that  the  adjacent  angles  formed  are  equal, 
each  of  these  angles  is  a  Right  Angle ;  as, 
Z  1  and  Z  2. 


27.  It  will  be  assumed  as  apparent  that    ^ ^  ^ 

all  right  angles  are  equal. 

28.  An  angle  is  measured  by  finding  how  many  times  it 
contains  another  angle  selected  as  unit  of  measure. 

The  usual  unit  of  measure  is  the  Degree,  which  is  one  ninetieth  of  a 
right  angle. 

To  express  fractional  parts  of  the  unit,  the  degree  is  divided  into  sixty 
equal  parts,  called  minutes,  and  the  minute  into  sixty  equal  parts,  called 
seconds. 

Degrees,  minutes,  and  seconds  are  represented  by  the  symbols  °,  ', 
and  "  respectively. 

Thus,  42°  22'  37"  denotes  an  angle  of  42  degrees,  22  minutes,  and  37 
seconds. 

Ex.  48.     Point  out  in  your  classroom  some  right  angles. 

Ex.  49.  Fold  a  piece  of  paper  so  as  to  make  a  straight  line  ;  then  fold 
it  again  so  as  to  form  two  equal  adjacent  angles.     Then  open  it  out. 

What  kind  of  angles  are  formed  by  the  lines  along  which  the  paper 
was  folded  ? 


Ex.  50.     How  many  degrees  are  there  in  : 
i  rt.  Z  ?     ^  rt.  Z  ?    1  rt.  Z  ?     |  rt.  Z  ? 

29.  Two  lines  are  Perpendicular  (±)   if 
they  form  a  right  angle.  C" 

Thus,  5^  ±  CZ)  if  Z  1  =  Z  2. 

When  two  lines  are  perpendicular,  the  ad- 
jacent angles  are  equal  (§  26). 

A  practical  method  of  drawing  a  perpendicular  to 
a  line  at  a  point  in  the  line,  is  to  use  a  pattern  right 
angle  as  in  the  adjoining  figure.  Draughtsmen  use 
the  right  triangle  illustrated.  A  card  with  two  per- 
pendicular edges  may  be  used  as  well. 


B 


14 


PLANE   GEOMETRY 


Ex.  51.  Draw  a  perpendicular  to  a  line  at  a  point  in  the  line  as 
suggested  in  §  29.  Use  a  card  or  else  the  pattern  right  angle  con- 
structed in  Exercise  49. 

Ex.  52.  Draw  a  perpendicular  to  a  line  from  a  point  not  in  the  line, 
using  a  pattern  right  angle. 

Ex.  53.  Draw  freehand  a  perpendicular  to  a  line  at  a  point  in  the 
line.  Test  the  accuracy  of  your  construction  by  means  of  your  pattern 
right  angle.  (Continue  this  exercise  until  you  can  draw  a  line  which  is 
approximately  perpendicular  to  a  given  line,  either  at  a  point  in  the  line, 
or  from  a  point  not  in  the  line.)  y^ 


30.   An  Acute  Angle  is  an  angle  which  is 
less  than  a  right  angle ;  as  Z  CBA. 


An  Obtuse  Angle  is  an  angle  which  is  greater 
than  a  right  angle  ;  as  Z  FEB. 

Acute   and  obtuse  angles   are  called  collectively 
Oblique  Angles. 

Two  intersecting  lines  which  are  not  perpendicular  are  said  to  be 
oblique  to  each  other. 

Ex.  54.  What  kind  of  angle  isZl?  Z2?  Z3? 
Z4  ?     (Test  each  with  your  pattern  right  angle.) 

Ex.  55.  What  kind  of  angle  do  the  hands  of  a 
clock  form  at  3  o'clock  ?  At  1  o'clock  ?  At  2 
o'clock  ?     At  5  o'clock  ? 

Ex.  56.  How  many  degrees  are  there  in  each 
of  the  angles  in  Exercise  55  ? 

31.  A  Straight  Angle  is  an  angle 
whose  sides  lie  in  a  straight  line  on 
opposite  sides  of  its  vertex.  ^ 

Such  an  angle  would  result  if  a  line  were  to  start  from  the  position  of 
line  OA  and  revolve  about  point  0  one  half  of  a  complete  revolution. 

32.  If  AB  is  any  straight  line,  and 
C01.AB,  then  Z  1  and  Z2  are  each 
right  angles  by  §  29 ;  also,  Z  AOB  is  a 
straight  angle  by  §  31.  Hence  a  straight 
angle  equals  two  right  angles. 


INTRODUCTION 


15 


33.  Since  a  straight  angle  is  equal  to  two  right  angles 
(§  32)  and  since  all  right  angles  are  equal  (§  27),  it  is  evident 
that  all  straight  angles  are  equal. 

Ex.  57.     How  many  degrees  are  there  in  a  straight  angle  ? 

Ex.  58.     What  part  of  a  straight  angle  is  a  right  angle  ? 

Ex.  59.    At  what  hour  do  the  hands  of  a  clock  form  a  straight  angle  ? 

34.  The  sum  of  all  the  successive  adjacent  angles  around  a 
point  on  one  side  of  a  straight  line  is  one  straight  angle. 

Thus,  Zl  +  Z2  +  Z3+Z4  =  ZAOB=  1  stZ. 

Ex.60.     If  Zl=Z2=Z3=Z4,howmany 
degrees  are  there  in  each  angle  ? 

.Ex.61.     If  Z2=3  times  Z1,Z3=Z2,  and  2? 
Z4  =  2  times  Zl,  how  many  degrees  are  there 
in  each  angle  ?     (Use  algebraic  method. ) 

35.  TTie    sum    of   all    the  successive 
adjacent  angles  around  a  point  is  two    jy 
sfraight  angles. 

What  kind  of  angle  is  Z 1  ?    Z2?     Hence  Zl +  Z2  =  ? 

The  total  angular  magnitude  around  a  point  is  called  a 
Perigon. 

Perigon  is  from  Greek  words  meaning  "  the  angle  around." 

Ex.  62.  How  many  right  angles  are  there  in  a  perigon  ?  How  many 
degrees  ? 

Ex.  63.  Through  what  angle  does  the  minute  hand  of  a  clock  turn  in 
one  half  hour  ?    In  one  hour  ? 

Ex.  64.  How  large  would  each  angle  of  the  adjoin- 
ing figure  be  if  the  angles  were  equal  angles  ? 

Ex.  65.  How  large  would  each  angle  of  the  adjoin- 
ing figure  be  if  Z 1  were  equal  to  Z  4,  if  Z  3  =  2  times 
Z  1,  if  Z5  =  the  sum  of  Z 3  and  Z 1,  and  if  Z2  =  the 
sum  of  Z3  andZ5? 

36.  Two  angles  are  Complementary  if  their 
sum  is  equal  to  a  right  angle.  Each  of  the 
angles  is  called  the  Complement  of  the  other. 

Thus,  the  complement  of  40^  is  60°. 


c 

^^ 

0 

A 

16 


PLANE   GEOMETRY 


Ex.  66.     What  is  the  complement  of  10^ 
ofO°?    of  45°?    ofr/? 


?   of  25"?    of  50°?     of  90°? 


(Uj 


Ex.  67.     How  large  is  the  angle  which  equals  its  complement 
algebraic  method. ) 

Ex.  68.     How  large  is  the  angle  which  equals  four 
times  its  complement  ? 

Ex.  69.     If    OC±OA  and   OD  ±  OB,  and  if  Z2 
=  70°,  compare  Z  1  and  Z  3. 

Ex.  70.  Draw  any  acute  angle.  Through  the  vertex  draw  a  line 
which  will  form  with  one  side  of  the  angle  the  complement  of  the  angle. 

37.  Complements    of  the    same 
angle  or  of  equal  angles  are  equal. 

The  complement  of  w°  is  (90  —  m)° 
and  the  complement  of  n°  is  (90  —  n)°. 

If,  now,  m  =  n,  then  90  —  m  must  equal  90  —  n,  for  when  equals  are  sub- 
tracted from  equals  the  remainders  are  equal. 

38.  Two  angles  are  Supplementary  if  their  sum  is  equal  to  a 
straight  angle.  Each  of  the  angles  is  called  the  Supplement  of 
the  other. 


Thus,  an  angle  of  150°  is  the  supplement  of  an 
angle  of  30°. 


Ex.  71.     What  is  the  supplement  of  80°  ? 
of  0°  ?    of  ic°  ?    of  Sx°? 


of  60°?    of  100°?    of  90°? 


Ex.  72.     How  large  is  the  angle  which  equals  its  supplement  ? 

Ex.  73.  How  large  is  the  angle  which  is  nine  times  as  large  as  its 
supplement  ? 

Ex.  74.  Draw  any  angle  less  than  a  straight  angle.  Through  its 
vertex  draw  a  straight  line  which  will  form  with  one  side  of  the  angle  the 
supplement  of  the  angle. 

39.  If  two  adjacent  angles  have  their  exterior  sides  in  a 
straight  line,  they  are  supplementary.  /^ 

In  the  adjoining  figure,  Z 1  and  Z  2  are  adjacent       ^ y^ 


angles.     OG  and    OA  are   their  exterior  sides.     If  0 

OC  and  OA  lie  in  a  straight  line,  then  ZAOG  =  1  st.  Z.     But  Z 1  +  Z2 

=  ZAOC.     Hence,  Z 1  and  Z  2  are  supplementary  (§  38). 


INTRODUCTION  17 

Two   adjacent   angles   which   are   also   supplementary   are 
called  Supplementary- adjacent  Angles. 

40.  If  two  adjacent  angles  are  supplementaiyj  their  extenor 
sides  lie  in  a  straight  line.     (Fig.  of  §  39.) 

For,  if  Zl  +  Z2  =  1  St.  Z,  then  Z.AOC  must  be  a  straight  angle  and 
AC  must  be  a  straight  Hne. 

Ex.  75.     Draw  two  adjacent  angles  which  are  not  supplementary. 

41.  Supplements  of  the  same   angle  or  of  equal  angles  are 
equal. 

In  the  adjoining  figure,  let  Z  1  =  Z3,  and  let  AC 
and  jB  T  be  straight  lines.  Then  Z  2  is  the  supple- 
ment of  Z  1  and  Z  4  is  the  supplement  of  Z3 ;  that 
is,  Z  2  =  180  -  Z  1  and  Z  4  =  180  -  Z  3.  Clearly, 
then,  Z  2  must  equal  Z  4,  for,  when  equal  angles 
are  subtracted  from  equal  angles,  the  remainders 
are  etjual. 

Ex.  76.     If  AB  and  CD  are  straight  lines,  are 
Z  1  and  Z  2  supplementary  ?  Z  3  and  Z  4  ?   Why  ?     '^  XHT 

(See  §39.) 

Suppose  that  Z1=Z4.     Must  Z3  then  equal     ^        /3 /4^ 
Z2?     Why? 


Ex.  77.     In  the  adjoining  figure,  if  Z  1  equals  Z2, 
then  Z  3  must  equal  Z  4.     Prove  it. 


42.   Two  angles  are  called  Vertical  Angles  when  the  sides  of 
one  are  prolongations  of  the  sides  of  the  other. 

Thus,  Z  1  and  Z  2  are  vertical  angles.     (See  Fig.  Ex.  78.) 

Ex.  78.     If  Z 1  =  40"^,  how  many  degrees  are, 
there  in  Z  3  ? 

How  many  degrees  are  there  in  Z  2  +  Z  3  ?  "' 

How  many  degrees  are  there  in  Z  2  ? 
How  then  do  Z  2  and  Z 1  compare  ? 


18 


PLANE  GEOMETRY 


Ex.  79.  Draw  two  straight  lines  that  intersect.  Make  a  tracing  of 
one  of  the  angles  formed  and  compare  it  with  its  vertical  angle.  What 
do  you  conclude  must  be  true  about  the  vertical  angles  ? 

43.   The  Protractor  is  a  tool  for  measuring  angles. 


The  point  0  on  the  protractor  will  be  referred  to  as  the  center  of  the 
protractor. 


44.   Problem.     Measure  a  given  angle  A  VB. 

fmun 


Place  the  protractor- over  the  given  angle  so  that  its  center  is  on  the 
vertex,  F,  of  the  angle,  and  its  edge  is  on  VA.  Then  read  from  the  pro- 
tractor the  number  opposite  the  point  where  VB  crosses  the  outer  edge 
of  the  protractor.  This  gives  the  number  of  degrees  in  the  angle  AVB. 
Thus,  in  the  figure,  ZAVB  -  50°. 


INTRODUCTION 


19 


Ex.  80.  Make  on  paper  a  tracing  of  the 
adjoining  figure.  On  your  paper,  extend  rays 
OA,  OB,  OC,  OD,  and  OE  until  they  are 
about  3  in.  in  length ;  then  measure  : 

(a)  ZAOB;  {b)/iAOC;  (c)  ZBOE; 
(d)  Z  COD. 


45.   Problem.     Construct  an  angle  of  given  size  at  a  point  in 
a  given  line. 

Construct  an  angle  of  35°  at  P  in  AB. 


Place  the  protractor  with  its  center  on  P  and  its  edge  on  PB  as  in  the 
figure.  Then  place  a  point  B  on  the  paper  opposite  the  35"  mark  on  the 
protractor.  Remove  the  protractor  and  draw  the  line  PB.  Then  angle 
BPB  equals  35°. 

Ex.  81.     Construct  with  the  protractor  an  angle  of  : 

(a)  70°;  (h)  40°;  (c)  65°;  (d)  100°;  (e)  143°. 
Write  below  each  angle  whether  it  is  an  acute  or  an  obtuse  angle. 


46.  A  Field  Protractor.  The  figure  at  top  of  page  20  repre- 
sents a  simple  field  protractor  which  can  be  made  by  some  mem- 
ber of  the  class.  With  it  angles  can  be  measured  in  the  field 
and  thus  some  elementary  surveying  problems  can  be  solved. 

On  a  flat  board  about  20  inches  square,  draw  a  circle  of  diameter  10 
inches.     Divide  its  circumference  into  360  equal  parts. 


20 


PLANE  GEOMETRY 


Make  an  arm  which  may  swing  about  the  center  of  the  circle  as  pivot. 

Let  the  arm  have  upon  it  two  "  sights  "  directly  in  line  with  the  center  of 

the  circle.  At  the  end  of  the  arm  and  in  line  with  the  sights  place  a  pin. 
The  board  may  be  attached  to  the  end  of  a 
stake  about  4  feet  long,  or  better  to  a  tripod. 

This  instrument  can  be  used  to  measure 
angles  in  the  open  field. 

Thus,  to  measure  an  Z.  CAB^  place  the  in- 
strument over  point  A.  Make  the  board  stand 
level.  (An  inexpensive  level  would  be  a  great 
help.)  Holding  the  board  stationary,  sight 
first  at  point  O,   and  read  the  angle  on  the 

protractor  ;  then  sight  at  point  5,  and  note  the  angle  on  the  protractor. 

The  number  of  degrees  through  which  the  arm  is  turned  in  passing  from 

^O  to  AB  is  the  measure  of  angle  GAB. 

47.  Triangle  (A).  Three  points  which  do  not  lie  in  the 
same  straight  line  determine  three  segments. 
Thus,  A,  B,  and  C  determine  the  segments 
AB,  BO,  and  AC.  The  figure  formed  by 
these  three  segments  is  called  the  triangle 
ABC  (A  ABC).  A,  B,  and  C  are  the 
vertices  of  the  triangle;  AB,  BC,  and  ^lOare         . 

the  sides  of  the  triangle ;  Z  A,  A  B,  and  Z  C  are  the  angles  of 
the  triangle. 

The  sides  and  angles  of  the  triangle  are  called  the  parts  of 
the  triangle.  They  are  six  in  number.  Opposite  each  side 
there  is  an  angle,  and  opposite  each  angle  there  is  a  side. 
Thus,  Z  (7  is  opposite  side  AB. 

48.  Experimental  Geometry.  Many  facts  about  geometrical 
figures  can  be  discovered  by  careful  drawing,  measurement, 
and  observation. 

Ex,  82.  On  a  line  AB,  at  a  point  P,  draw  a  ray  PC  making  Z  APC 
=  80°.  Measure  Z  CPB.  What  fact  studied  previously  does  this  exer- 
cise verify  ? 

Ex.  83.  Draw  two  intersecting  straight  lines.  Measure  each  of  a 
pair  of  vertical  angles.  What  fact  studied  previously  does  this  exercise 
verify  ? 


INTRODUCTION  21 

Ex.  84.  Construct  a  A  ABC,  having  AB  and  BC  each  3  inches 
long  and  Z  B  =  40°.  Measure  Z  A  and  Z  C.  IIow  do  they  compare  ? 
(A  triangle  having  two  equal  sides  is  called  an  isosceles  triangle.) 

Ex.  85.  Draw  any  triangle  having  two  equal  sides.  Measure  the 
angles  opposite  the  equal  sides.  How  do  they  compare  ?  What  fact  is 
suggested  by  Exercises  84  and  85  ? 

Ex.  86.  Draw  any  triangle  of  reasonably  large  size.  Measure  each 
of  its  angles.  Find  their  sum.  Kepeat  the  exercise  for  another  triangle 
of  somewhat  different  shape.  Compare  your  results  with  those  of  some 
other  pupils.     What  seems  to  be  the  sum  of  the  angles  of  a  triangle  ? 

Ex.  87.  Draw  a  A  ABC  in  which  AB  and  BC  are  each  3  inches  and 
ZB  =  50°.  Let  E  be  the  mid-point  of  BC,  and  F  the  mid-point  of  AB. 
Draw  AE  and  CF.    Measure  them.     What  seems  to  be  true  ? 

Ex.  88.  Draw  any  triangle  ABC  of  reasonably  large  size.  Let  E  be  the 
mid-point  of  AB,  and  F  the  mid-point  of  BC.  Draw  EF.  Compare  EF 
and  AC  by  measuring  them.    What  seems  to  be  the  relation  between  them  ? 

Ex.  89.  Draw  a  segment  AB.  At  its  center,  C,  draw  a  line  CD 
perpendicular  to  AB.  From  E,  any  point  on  CD,  draw  AE  and  BE. 
Compare  them  by  means  of  your  dividers.  Take  any  other  point  on  CD 
and  repeat  the  exercise.     What  fact  seems  to  be  suggested  ? 

Ex.  90.  Let  AB  be  any  line  segment.  Draw  CA  ±  AB  at  A,  and 
DB  LAB  At  B.  Make  CA  -  DB.  Draw  AD  and  CB.  Compare  them 
either  by  measurement  or  by  means  of  the  dividers. 

49.  Objections  to  studying  geometry  only  hy  the  experi- 
mental method  may  be  given.  To  get  satisfactory  results,  the 
figures  must  be  drawn  and  measured  with  greater  accuracy  than 
is  usually  possible.  Conclusions  reached  from  the  study  of  one 
or  two  special  figures  may  be  incorrect.  Frequently  one  is  mis- 
led by  assuming  relations  ivhich  appear  to  the  eye  to  be  correct. 


Ex.  91.    In  the  first  figure  above,  are  ^B  and  CD  straight  lines  ? 
Ex.  92.     In  the  second  figure  above,  is  AB  equal  to  or  less  than  CD  ? 


22  PLANE   GEOMETRY 

50.  Demonstrative  Geometry.  For  the  reasons  given  in  §  49 
and  for  other  reasons,  it  is  customary  to  study  geometry  by 
what  is  known  as  the  demonstrative  method.  Statements  are 
not  accepted  until  they  are  proved  to  he  true,  except  for  a  few 
fundamental  ones  which  are  assumed  as  a  foundation. 

51.  An  Axiom  is  a  statement  accepted  as  true  without 
proof.  Usually  the  truth  is  very  evident.  The  following 
are  important  axioms ;  others  will  be  introduced  as  they  be- 
come necessary. 

AXIOMS 

Ax.  1,    Quantities  which  are  equal  to  the  same  quantity  or  to 

equal  quantities  are  equal  to  each  other.     (See  Ex.  22 

and  Ex.  23.) 
Ax.  2.    Any  quayitity  may  he   suhstituted  for   its  equal  in  a 

mathematical  expression. 
Ax.  3.    If  equals  he  added  to  equals,  the  sums  are  equal.     (See 

Ex.  24.) 
Ax.  4.    If  equals  he  suhtracted  from  equals,  the  remainders  are 

equal. 
Ax.  5.    If  equals   he   multiplied   hy  equals,  the  products  are 

equal. 
Ax.  6.   If  equals  he  divided  hy  equals,  the  quotients  are  equal. 

(The  divisor  must  not  be  zero.) 
Ax.  7.    The  whole  equals  the  sum  of  its  parts. 
Ax.  8.    TJie  whole  is  greater  than  any  of  its  parts. 
Ax.  9.    If  a  and  h  are  any  two  magnitudes  of  the  same  hind, 

then  a  is  less  than  h,  is  equal  to  h,  or  is  greater  than  h. 
Ax.  10.    Oyily  one   straight   line   can   he   draivn   through   two 

points.     (§  8.) 
Ax.  11.    The  straight  line  segment  is  the  shortest  line  that  ca7i 

he  drawn  hetween  two  points.     (§  14.) 
Ax.  12.   All  right  angles  are  equal.     (§  27.) 
Ax.  13.   An  angle  has  only  one  hisector.     (§  23.) 
Ax.  14.   A  segment  has  only  one  mid-point.     (§  15.) 


INTRODUCTION  23 

52.  A  Theorem  is  a  statement  which  requires  proof.  Every 
theorem  can  be  expressed  by  a  sentence  which  has  one  clause 
beginning  with  "  if "  and  a  second  clause  beginning  with 
"  then." 

The  clause  beginning  with  "if'  is  called  the  Hypothesis; 
it  indicates  what  is  known  or  assumed. 

The  clause  beginning  with  "  then  "  is  called  the  Conclusion ; 
it  states  what  is  to  be  proved. 

Thus:  (Hypothesis)     If  two  sides  of  a  triangle  are  equals 
(Conclusion)     then  the  angles  opposite  are  equal. 

53.  Some  theorems  have  been  proved  already  in  an  informal 
manner. 

INFORMALLY  PROVED  THEOREMS 

1.  Two  straight  lines  can  intersect  at  only  one  point,     (§  11.) 

2.  All  radii  of  the  same  circle  or  of  equal  circles  are  equal. 

(§  17.) 

3.  A  straight  angle  equals  two  right  angles.     (§  32.) 

4.  All  straight  angles  are  equal.     (§  33.) 

5.  The  sum  of  all  the  successive  adjacent  angles  around  a 

point  on  one  side  of  a  straight  line  is  one  straight  angle. 
(§  34.) 

6.  Tlie  sum  of  all  the  successive  adjacent  angles  around  a 

point  is  two  straight  angles.     (§  'do.) 

7.  Complements  of  the  same  angle  or  of  equal  angles  are 

equal.     (§  37.) 

8.  If  two  adjacent  angles  have  their  exterior  sides  in  a  straight 

line  J  they  are  supplementary.     (§  39.) 

9.  If  two  adjacent  angles  are  supplementary,  their  exterior 

sides  lie  in  a  straight  line.     (§  40.) 
10.   Supplements  of  the  same  angle  or  of  eqvM  angles  are  equal. 
(§  41.) 

54.  In  a  formal  demonstration  or  proof,  each  statement 
made  is  proved  by  quoting  a  definition,  an  axiom,  the  hypothe- 
sis, or  some  previously  proved  theorem. 


24 


PLANE   GEOMETRY 


ILLUSTRATIVE  DEMONSTRATION 

Theorem.    If  two  straight  lijies  intersect,  the  vertical 
angles  are  equal. 


Hypothesis.     St.  lines  AB  and  CD  intersect  at  0, 
forming  vertical  Al  and  2. 

Conclusion.  Z  1  =  Z  2. 

Proof.     1.  AB  is  a  straight  line. 

2.  .*.  Z  1  is  a  supplement  of  Z  3. 

[If  two  adj.  A  have  their  ext.  sides  in  a  st.  line, 
they  are  supplementary.] 


Hyp. 
§39 

Hyp. 
§39 
§  41 


3.  CD  is  a  straight  line. 

4.  .*.  Z  2  is  a  supplement  of  Z  3. 

5.  .-.  Z1  =  Z2. 

[Supplements  of  the  same  Z  are  equal.] 

Note.  — This  theorem  was  apparently  assumed  by  Thales. 

Ex.  93.     Prove  in  the  same  manner  that  Z  AOD  =  Z  COB. 

Ex.  94.     If  Z  3  =  130°,  how  many  degrees  are  there  in  each  of  the 
other  angles  in  the  figure  above  ? 

Ex.  95.     Two  lines  intersect  so  that  one  of  the  angles  is  a  right  angle. 
How  large  is  each  of  the  other  angles  formed  ? 

55.   Besides  the  proofs  of  certain  theorems,  the  methods  of 
constructing  certain  figures  are  studied  in  geometry. 

A  Problem  is  a  construction  to  be  made. 


INTRODUCTION  25 

56.  A  Postulate  is  a  construction  assumed  possible. 

The  following  postulates  are  necessary  at  the  present  time. 

POSTULATES 

1.  One  straight  line  can  be  drawn  through  two  points.     (§  8.) 

2.  A  straiqht  line  can  he  extended  indefinitely  in  each  direction. 

(§  8.) 

3.  A  circle  can  be  drawn  with  any  point  as  center  and  any 

given  segment  as  radius. 

57.  The  word  Proposition  is  commonly  used  to  designate  a 
theorem  or  a  problem  discussed  in  the  text. 

SUPPLEMENTARY  EXERCISES 

Ex.  96.  Draw  a  line  to  represent  the  path  of  a  baseball  when  the 
pitcher  throws  an  "out-curve." 

Ex.  97.  A  farmer  is  setting  out  trees  for  an  orchard.  He  first  places 
the  trees  which  are  at  the  ends  of  a  row.  How  may  he  then  locate  the 
other  trees  of  that  row  so  that  they  will  be  in  a  straight  line  without  using 
a  long  line  between  the  two  end  trees  ?  What  two  points  determine  the 
straight  line  formed  by  the  trees  ? 

Ex.  98.  How  do  plasterers  use  in  their  work  the  characteristic 
property  of  a  plane  mentioned  in  §  1  ?  200^ 

Ex.  99.  A  man  wishes  a  scale  drawing  of 
his  suburban  lot  so  that  he  may  consult  a  land- 
scape architect  about  the  proper  planting  of  it. 
He  made  the  adjoining  rough  drawing  of  the  lot, 
and  then  obtained  the  measurements  indicated. 
Make  a  scale  drawing  of  the  lot,  letting  J  inch 
represent  25  feet.  „^_ 

Supplementary  and  Complementary  Angles 

Ex.  100.     What  is  the  complement  of  40°  30'  ? 

Ex.  101.     What  is  the  supplement  of  56^  30'  ? 

Ex.  102.  How  many  degrees  are  there  in  an  angle  if  its  complement 
contains  40°  ? 


26  PLANE   GEOMETRY 

Ex.  103.     How  many  degrees  are  there  in  an  angle  whose  supple- 
ment contains  80°  ? 

Ex.  104.    Find  the  angle  whose  supplement  is  ten  times  its  comple- 
ment. 

Ex.  105.     Two  angles  are  complementary.    The  greater  exceeds  the 
less  by  25°.     Find  the  angles.     (Use  the  algebraic  method.) 

Ex.  106.     Find  the  angle  which  exceeds  its  supplement  by  34°. 

Ex.  107.     The  sum  of  the  supplement  and  complement  of  a  certain 
angle  is  140°.    Find  the  angle. 

Ex.  108.     Find  the  number  of  degrees  in  the  angle  the  sum  of  whose 
supplement  and  complement  is  196°. 

Ex.  109.     The  supplement  of  a  certain  angle  exceeds  three  times 
its  complement  by  18°.     Find  the  angle. 

Ex.  110.     Prove  that  the  supplement  of  any  angle  exceeds  its  com- 
plement by  one  right  angle. 

Vertical  Angles 

Ex.  111.    Prove  that    the    straight    line  which 
bisects  an  angle  also  bisects  the  vertical  angle. 
Hyp.     OE  bisects  AAOC.    j^ Oi^  is  a  st.  Ihie. 
Con.      OF  bisects  Z.  BOD. 

Ex.  112.     Prove  that  the  bisectors  of  two  supplementary  adjacent 
angles  are  perpendicular.  .E  /D 


Suggestions.  — L1  =  \LBCD;  L2^h^DCA.  \       A     ^^ 

Zl  +  Z2  =  ?  ^  ^  ^' 

A ^ B 

Ex.  113.     If  the  bisectors  of  two  adjacent  angles  are  perpendicular, 
the  angles  are  supplementary.     (See  figure  of  Ex.  112.) 

Ex.  114.     If  the  bisectors  of  two  adjacent  angles  make  an  angle  of 
45°,  the  angles  are  complementary 


Ex.115.     Hyp.     CO  bisects  2^:^05.     DEA.CO. 
Con.     Z3  =  Z4. 

Suggestions.  —  1.  Compare  Z 1  and  Z.  2. 

2.  Compare  L  3  and  Z 1 ;  Z  4  and  Z  2. 


INTRODUCTION 


27 


Ex.  116.     Hyp.    A  ABC  is  a  rt.  Z. 

Z  3  is  complementary  to  Z.  1. 
Con.     Z3  =  Z2. 


D  *rr" 


Ex.  117.'    Two  straight  lines  intersect  so  that  one  angle  formed  is 
60°.     How  large  is  each  of  the  other  angles  ? 

Ex.  118.     Review  the  following  definitions : 

(a)  Segment  of  a  straight  line.  Qi)  Equal  angles. 


(6)  Equal  segments. 

(c)  Mid-point  of  a  segment. 

(d)  Circle. 

(e)  Radius  and  diameter. 
(/)  Ray  or  half-line. 

(^)  Angle  ;  vertex  ;  side. 


(i)  Bisector  of  an  angle. 

(j)  Right  angle  ;  acute  ;  obtuse  ; 
straight. 

{Ic)  Complementary  angles  ;  supple- 
mentary ;  adjacent ;  vertical. 

(Z)  Perpendicular  lines. 


Ex.  119.     What  is  an  axiom  ?  a  theorem  ? 

Ex.  120.     What  is  the  hypothesis  of  a  theorem  ? 
elusion  of  a  theorem  ? 


What  is  the  con- 


Ex.  121.  Lay  off  on  a  field  some  irregular  piece  of  ground.  Obtain 
such  measurements  as  will  enable  you  to  make  a  scale  drawing  of  the 
field.     (This  exercise  is  similar  to  Ex.  99.) 

Ex.  122.  Another  method  for  ob- 
taining measurements  for  a  scale 
drawing  for  a  piece  of  ground  and 
objects  upon  it  is  to  locate  the  instru- 
ment for  measuring  angles  at  a  point 
like  0  in  the  adjoining  figure.  Then 
find  the  distances  of  the  other  points 
from  0  and  their  directions  from  OB 
or  from  OA. 

Select  an  irregular  piece  of  ground  and  obtain  the  measurements  which 
will  enable  you  to  make  a  scale  drawing  of  it  and  locate  upon  the  draw- 
ing some  of  the  trees  or  other  objects  on  the  lot. 


Supplementary  Notes  on  Definitions 

Note  1.  —  Toird  and  straight  line  are  undefined.     (See  §4  and  §5.) 
A  definition  describes  a  term  by  means  of  simpler  terms.     It  is  evident 


28  PLANE   GEOMETRY 

then  that  there  must  be  some  terms  which  cannot  be  defined,  as  there  are 
no  terms  simpler  than  them  by  which  to  define  them. 

No  definition  of  point  can  be  given. 

No  satisfactory  definition  of  straight  line  suitable  for  high  school  pupils 
can  be  given.  n 

Hence  point  and  straight  line  are  left  undefined.  y^ 

Note  2. — Belating  to  the  definition  of  an  angle           f       /y^ 
(§20).  ^^0-^ A 

Two  rays  OA  and  OB  actually  form  two  angles ;  V^_^ 

namely,  Z  1  and  Z2  of  the  adjoining  figure.  In  Z  1,  OA  is  the  initial  line 
(§  20)  and  OB  is  the  terminal  line  ;  in  Z2,  OB  is  the  initial  line  and  OA 
is  the  terminal  line.  Usually,  one  angle  is  less  than  and  the  other  is 
greater  than  a  straight  angle. 

Unless  something  is  said  to  the  contrary,  Z  J. 0J5  refers  to  the  smaller 
angle  formed  by  the  rays  OA  and  OB. 


BOOK  I 

KECTILINEAR  FIGURES 

58.  A  Rectilinear  Figure  is  a  geometrical  figure  composed 
of  straight  lines  only. 

59.  Congruence.  If  asked  to  compare  two  sheets  of  paper 
as  to  shape  and  size,  it  is  natural  to  place  one  upon  the  other  to 
determine  whether  they  can  be  made  to  coincide  (fit  together). 

60.  Two  geometrical  figures  are  congruent  (^)  if  they  can  be 
made  to  coincide. 

61.  Ax.  15.  Congruence  Axiom.  —  Two  figures  which  are 
congruent  to  the  same  figure  are  congruerit  to  each  other. 

Historical  Note.  —  The  symbol  ^  was  introduced  by  a  mathemati- 
♦cian,  Leibnitz,  in  1679. 

62.  Superposition  is  the  process  of  placing  one  geometrical 
figure  upon  another  for  the  purpose  of  comparing  them. 
Literally,  superpose  means  "  place  above." 

Postulate.  —  A  geometrical  figure  may  be  moved  about  in  space 
icithout  changing  any  of  its  parts. 

Ex.  1.  Notice  the  panes  of  glass  in  the  windows  of  your  schoolroom. 
Do  they  appear  to  be  congruent  ?     Do  they  coincide  now  ? 

Ex.  2.     Draw  A  ylBC,  having  ^5  =  4  in.,  BC  =  Q  in.,  and  Zi?  =  50°. 

(a)  Measure  ^A,  ZC^  and  AC. 

(h)  Cut  your  triangle  from  the  paper.  Compare  it  by  superposition 
with  the  triangles  made  by  other  members  of  your  class. 

(c)  What  do  you  conclude  must  be  true  about  all  triangles  made 
according  to  the  directions  ? 

Ex.  3.  Are  the  statements  of  the  hypothesis  assumed  to  be  true  or 
must  they  be  proved  to  be  true  ?  Answer  the  same  question  for  the 
conclusion. 

29 


30  PLANE   GEOMETRY  — BOOK  I 

Proposition  I.     Theorem 
63.   If  hvo  triangles  have  two  sides  and  the  included 
angle  of  one  equal  respectively  to  two  sides  and  the 
included  angle  of  the  other,  the  triangles  are  congruent. 


B  B  E 

Hypothesis.     In  A  ABC  and  A  DBF: 

AB  =  DE;  AC=DF;  ZA  =  ZD. 

Conclusion.  A  ABC  ^  A  DEF. 

Proof.     1.   Place  A  ABC  on  A  DEF  with  point  A  on  point 
D,  and  side  AB  on  side  DE. 

2.  Point  B  will  fall  on  point  E. 

[Since  AB  =  DE,  by  hypothesis.]  §  13 

3.  AC  will  fall  on  DF. 

[Since  ZA  =  ZD,  by  hypothesis.]  §  21 

4  Point  C  will  fall  on  point  F. 

[Since  AC  =  DF,  by  hypothesis.]  §  13 

5.  .-.  BC  must  coincide  with  EF. 

[Only  one  st.  line  can  be  drawn  through  two  points.]     Ax.  10  ;  §  51 

6.  .-.  A  ABC  coincides  with  A  DEF  and  is  congruent  to  it. 

[Two  A  are  congruent  if  they  can  be  made  to  coincide.]  §  60 

Ex.  4.     After  you  place  A  ABC  so  that  A  falls  on  D,  does  AB  fall  on 
DE,  or  must  you  place  AB  on  DE  ? 

Ex.  5.     Where  would  C  fall  if  AC  were  equal  to  |  DF?     Would  the 
triangles  be  congruent  ? 

Ex.  6.     Where  would  AC  fall  'd  ZA  were  less  than  ZD? 

Ex.  7.     Make  a  free-hand  drawing  to  illustrate  the  result  if  a  A  ABC 
is  superposed  on  a  A  DEF,  when  AC=i  DF,  ZA=ZD,  and  AB=^  DE. 


RECTILINEAR  FIGURES 


31 


64.  Application  of  First  Theorem.     Illustrative  Exercise. 


Hypothesis. 

AB^CD. 

i\ 

Z  1  =  Z  2. 

^^o>^ 

Conclusion. 

A  ABC  ^  A  BCD. 

^^O^ 

Proof.     1. 

In  A  ABC  and  A  BCD: 

J) 

AB=CD', 

Hyp. 

Z1  =  Z2; 

Hyp. 

BC  =  BC. 

See  note  below. 

2. 

.'^A  ABC  ^  A  BCD, 

[If  two  A  have  two  sides  and  the  included  Z  of  one 
equal  respectively  to  two  sides  and  the  included  Z  of  the 
other,  the  A  are  congruent.] 


Note.  — The  symbol  =  is  read.  "  is  identically  equal  to." 
thority  is  required,  for  any  magnitude  is  equal  to  itself. 

Ex.  8.     Hyp.  Zl=Z2;  AB  =  EC. 

Con.  AABD^ABCD.  ' 


No  other  au- 


■^' 


Ex.  9.     Hyp. 


Con. 


Suggestion. 
Why? 


AD  and  BC  are  st.  lines. 
A0=  OD]  B0=  OC. 
AABO^ACDO. 

What  do  you  know  about  Zl   and   Z2? 


Ex.  10.     Hyp. 

Con. 

Ex.  11.     Hyp. 

Con. 


XY  =  XW. 
XZ  bisects  Z  X. 

AXYZ^AXZW. 

DBC  is  a  St.  line. 
ABA. DC)  DB  =  BC. 
A  ADB  ^  A  ABC. 


65.   Homologous  parts  of  congruent  figures  are  parts  which 
are  similarly  located ;  they  are  the  parts  which  coincide  when 


32 


PLANE    GEOMETRY  — BOOK  I 


the  figures  are  made  to  coincide.     It  follows  that :  homologous 
parts  of  congruent  figures  are  equal. 

Thus,  in  §  63,  Z  O  is  homologous  to  Z  F^  and  side  J5(7  is  homologous 
to  side  EF.     It  follows  that  AC  =  A  F  and  BC  =  EF. 

Note.  —  In  congruent  triangles,  homologous  sides  lie  opposite  equal 
angles  and  homologous  angles  lie  opposite  equal  sides. 

Q6.  Principle  I.  To  prove  that  two  segments  are  equal  or 
two  angles  are  equal,  try  to  prove  them  homologous  parts  of 
congruent  triangles. 

Illustrative  Exercise 


Hyp.    BG  is  a  st.  line.                          j 
AB±BC',  DC±BG; 

\ 

AB=DC. 

1      ^\o 

rf 

0  is  mid-point  of  BC. 

2 

\-/ 

Con.     AO  =  OD. 

\J 

T) 

Plan.    Try  to  prove  AO  and  OD 
homologous  sides  of  congruent  A. 

Proof.     1.        Since  ABA.BC,  Al  =  2^  rt.  Z. 

2.  Since  DC  1.  BC,Z2  =  ^  rt.  Z. 

3.  .-.  Z1=Z2. 

Def. 
Def. 

[All  rt,  Zs  are  equal.] 

4.  In  A^jBOand  A7)C0: 

AB^DC;  Hyp: 

Z1  =  Z2;  Step  3 

BO=OC.  Hyp. 

5.  .'.  AABO^ADCO. 

[If  two  A  have  two  sides  and  the  included  Z  of  one  equal 
respectively  to  two  sides  and  the  included  Z  of  the  other,  the 
A  are  congruent.] 

6.  .'.AO=OD. 

[Homologous  sides  of  cong.  A  are  equal.] 

Note.  —  AO  lies  opposite  Z  1  and  OD  lies  opposite  Z2;  hence  they 
are  homologous  sides  of  the  congruent  triangles. 


RECTILINEAR  FIGURES 

Ex.  12.    Hyp.  AB±BC;  DC±BC; 

O  bisects  BC;  AB  =  DC. 
Con.  A0=  OD. 

Ex.  13.   Hyp.  NO  bisects  Z  PNM. 
MN=NP. 
Con.  Zilf  =  ZP. 


Ex.  14.   Hyp.   AB  =  BC=  CD 
ZB  =  ZD. 
Con.  AC  =  CE. 


DE. 


Ex.  15.   To  obtain  tlie  distance  AB. 

(1)  Locate  point  O  from  which  OA  and  OB  may  be 
measured.  (2)  Extend  AO  and  50,  making  OC=AO 
and  0D  =  BO.    Then  DO  =  ^i?.     Prove  it. 


Ex.  16.    If  AB  and    CD  are  two  diameters  of  a 
circle,  prove  that  AD  must  equal  BC. 


Review  Exercises 

Ex.  17.    If,  in  the  adjoining  figure,  Z  3  =  Z  7, 
prove  Z  2  =  Z  7.  ~ 

Ex.  18.    If  Z  4  =  Z  5,  prove  Z  1  =  Z  8.  ^~ 

Ex.  19.   If  Z  3  =  Z  G,  prove  Z  1  =  Z  5.  ^' 

Suf^gestioiis.  —1.  Recall  §41.    2.  Of  what  angle  is  Zl  the  supplement? 
3.  Of  what  angle  is  Z 5  the  supplement? 

Ex.20.    If  Z4  =  Z8,  proveZ3  =  Z6. 

Ex.  21.    When  are  two  figures  congruent  ? 

Ex.  22.   What  method  of  proof  is  employed  in  proving  Proposition  I  ? 


Ex.  23.  Draw  a  A ^50,  having  ^B=4  in.,  ZA  =  60°,  and  Z  5=80°. 
Cut  your  triangle  from  the  paper  and  compare  it  with  the  triangles  made 
by  other  members  of  your  class.  What  do  you  conclude  must  be  true 
about  all  triangles  made  according  to  the  directions  given  ? 


34  PLANE  GEOMETRY  — BOOK  I 

Pkoposition  II.     Theorem 

67.   If  two  triangles  have  two  angles  and  the  included 
side  of  one  equal  respectively  to  two  angles  and  the  irv- 
eluded  side  of  the  other,  the  triangles  are  congruent 
c 


A  B  D 

Hypothesis.  In  A  ABO  and  A  DEF: 

AB  =  DE;  ZA=ZD;  Z.B=ZE. 

Conclusion.  A  ABC  ^  A  DEF. 

Proof.  1.  Place  A  ABC  on  A  DEF,  with  point  A  on  point 
D  and  AB  on  DE. 

2.  Point  B  will  fall  on  point  E. 

[Since  AB  =  BE,  by  hypothesis.]  §  13 

3.  AC  will  fall  on  DF,  C  falling  somewhere  on  line  DF. 

[Since  ZA=ZD,\)y  hypothesis.  ]  §  21 

4.  BC  will  fall  on  EF,  0  falling  somewhere  on  line  EF. 

[Since  /.B  =  AE^hy  hypothesis.]  §  21 

5.  .*.  point  C  must  fall  on  point  F. 

[Two  St.  lines  can  intersect  at  only  one  point.]  §  11 

6.  .-.  A  ABC  coincides  with  AZ>^Fandis  congruent  to  it. 

[Two  A  are  congruent  if  they  can  be  made  to  coincide.  ]  §  60 

Ex.  24.    Where  would  AC  fall  if  Z  A,  above,  were  less  than  /.D'> 

Ex.  25.  Draw  freehand  the  approximate  figure  which  would  result 
from  superposing  A  ABC  on  A  DEF  if  AB  =  DE,  AA  =  ^ZD,  and 
ZB  =  iZE. 

Ex.  26.  After  proving  A  ABC  congruent  to  A  DEF,  what  do  you 
know  about :  (a)ZC?     Why ?     (6)  About  AC?     (c)  About  BC?    § 65 


RECTILINEAR  FIGURES  35 

Ex.  27.   In  Propositions  I  and  II,  how  many  parts  (§  47)  of  one  triangle 
are  given  equal  to  parts  of  the  other  triangle  ?  ^ 

Ex.28.   Hyp.    Z1=Z2.  /^^"^>^ 

Con.    A  ABC  ^  A  BCD.  ^^^^ 

Ex.  29.   Hyp.   AE  and  BD  are  st.  lines. 
ZB=  ZD:  C  bisects  BD. 


t^ 


Con.  ZA  =  Z£.  ^^^ 

Suggestion.  —  Read  §  66.  ^ 

Ex.30.   Hyp.    ^C  bisects  Z  a  /, 

BG±AD. 
Con.   AB  =  BD. 


Ex.  31.  If  the  line  which  joins  two  opposite  vertices  of  a  quadrilateral 
(four-sided  figure)  bisects  the  angles  whose  vertices  it  joins,  then  the 
other  two  angles  are  equal.  b^ 

Hyp.    Z1  =  Z2;  Z3  =  Z4. 
Con.   ZB  =  ZD. 

D 

Note.  —  Supplementary  Exercises  1  and  2,  p.  273,  can  be  studied  now. 

68.  A  triangle  is  Scalene  when  no  two  of  its  sides  are  equal ; 
Isosceles  when  two  of  its  sides  are  equal ;  Equilateral  when  all 
its  sides  are  equal ;  Equiangular  when  all  its  angles  are  equal. 


Scalene  Isosceles  Equilateral 

A  triangle  can  be  made  to  "  stand  upon  "  any  one  of  its  sides. 
Hence  any  side  of  a  triangle  can  be  considered  its  Base.  When 
a  side  has  been  selected  as  base,  the  opposite  vertex  is  called 
the  Vertex  of  the  triangle,  and  the  angle  at  that  vertex  is  called 
the  Vertical  Angle  of  the  triangle. 

In  an  iso.sceles  triangle,  the  side  which  is  not  one  of  the 
equal  sides  is  usually  taken  as  the  base ;  and  then  the  angle 
formed  by  the  equal  sides  is  the  vertical  angle  of  the  isosceles 
triangle. 


36 


PLANE  GEOMETRY  — BOOK  I 


Pkoposition  III.     Theorem 

69-   In  a7i  isosceles  triangle,  the  angles  opposite  the 
equal  sides  are  equal. 


Proof. 

2. 


B 

Hypothesis.  In  A  ABC,  AC=  BC. 

Conclusion.  /.A  =  /.B. 

Plan.     Prove  ZA  and  Z  B  homologous  A  of  cong.  A. 
1.  Let  CD  bisect  Z  O. 

In  A  ACDsLTid  A  BCD: 

AC=BC;  Hyp. 

CD  =  CD.  See  Note  1,  §  64. 

Z  1  =  Z  2.  Def . 

[Since  CD  bisects  Z  O.  ] 
r.AACD^ABCD. 

[If  two  /^  have  two  sides  and  the  included  Z  of  one 
equal  respectively  to  two  sides  and  the  included  Z  of  the 
other,  the  A  are  congruent.  ] 

.'.ZA  =  ZB. 

[Homologous  A  of  cong.  ii  are  equal.] 

Note  1.  —  Principle  I  (§  66)  is  used.  To  get  two  triangles,  a  construc- 
tion line,  CD,  was  drawn.     This  is  often  necessary. 

Note  2.  —  This  theorem  is  ascribed  to  Thales,  although  this  may  not 
be  his  proof.  The  proof  for  this  theorem  which  was  given  by  Euclid  ap- 
pears as  Ex.  3,  p.  273.     It  can  be  studied  after  §  73. 

Ex.  32.  Why  are  ZA  and  ZB  homologous  angles  of  the  congruent 
triangles  in  the  proof  of  Proposition  III  ? 

70.     Cor.    An  equilateral  triangle  is  also  equiangular. 
(Read  §  71  at  this  time.) 


3. 


4. 


63 


§65 


RECTILINEAR  FIGURES 


37 


71.  A  Corollary  is  a  theorem  which  is  easily  deduced  from 
the  theorem  with  which  it  is  given.  For  each  corollary,  draw 
a  figure,  form  the  hypothesis  and  conclusion,  and  give  the 
proof. 

Ex.  33.    Hyp.         AB  =  AC. 
Zl  =Z2. 
Con.  AABX^AAYC. 
Suggestion.  — Does  Z.3  =  Z'^'!    Why? 

Ex.  34.    If  AB  =  AC  and  if  D  is  the  mid-point 
of  BC,  Eoi  AB,  and  F  of  AC,  prove  that  ED=FD. 

Suggestions.  — Form  the  hypothesis  and  conclusion. 
Read  §  (56.    Does  ^A'  =  CF  ?    Why  ? 

Ex.  35.  After  proving  DE  =DF  in  Ex.  34,  draw  EF  and  prove  that 
ZDEF=ZDFE. 

Ex.  36.    In  the  figure  drawn  for  Ex.  35,  prove  that  Z  AFE  =  Z  AEF. 

Ex.  37.  If  AB  and  CB  are  two  rafters  of 
equal  length  in  a  roof,  and  if  DF  and  EG  are 
supports,  perpendicular  to  the  floor  AG,  at 
points  equally  distant  from  A  and  C  respec- 
tively, prove  that  DF  must  equal  EG.  (Form 
the  hypothesis  and  conclusion  first. ) 

Ex.  38.  In  an  isosceles  triangle  ABC,  having  AB  =  AC,  point  D  is 
any  point  in  the  base  BC.  E  ia  taken  on  AC  and  F  on  AB  so  that  EC 
=  BD  and  BF  =  DC.  Prove  that  DE  =  DF.  (Draw  the  figure  as  it  is 
described.) 


Ex.  39.  To  obtain  the  distance  AB. 

(1)  LayoffjBCJLto^^.  (2)  Lay  off  (7jE:±  to  ^O. 
(3)  Place  a  stake  at  0,  the  mid-point  of  BC.  (4)  De- 
termine, by  sighting,  a  point  D  on  CE  so  that  A,  O, 
and  D  will  be  in  the  same  straight  line.  Then  CD 
=  AB.     Prove  it. 

Ex.  40.   To  obtain  the  distance  AB. 

(1)  Let  AC  he  any  convenient  segment.  (2)  Lay 
off  AD,  making  ZS  =  Zl.  (3)  Lay  off  CD,  making 
Z  4  =  Z  2.     Then  AB  =  AD.     Prove  it. 


38  PLANE   GEOMETRY  — BOOK  I 

Proposition  IV.     Problem 
72.    Construct  a  triangle,  having  given  its  three  sides. 


Given  m,  n,  and  p,  the  three  sides  of  a  triangle. 
Required  to  construct  the  triangle. 
Construction.     1.   Draw  AB  =  m. 

2.  With  A  as  center,  and  n  as  radius,  draw  an  arc. 

3.  With  B  as  center,  and  p  as  radius,  draw  a  second  arc,  in- 
tersecting the  first  arc  at  C. 

4.  Draw  AC  and  BO. 

Statement.  A  ABC  is  the  required  triangle,  as  it  has  the 
given  sides. 

Discussion.  If  one  side  is  equal  to  or  greater  than  the  sum 
of  the  other  two  sides,  the  construction  is  impossible. 

Ex.  41.  A  piece  of  ground  is  triangular  in  form.  Its  sides  measure 
100  rd.,  150  rd.,  and  200  rd.,  respectively.  Make  a  scale  drawing  of  the 
triangle,  letting  1  in.  represent  100  rd. 

Ex.  42.  Construct  an  isosceles  triangle  whose  base  is  2  in.  and  whose 
equal  sides  are  each  3  in. 

Ex.  43.  A  girl  wants  an  equilateral  triangle  whose  sides  are  each  3  in. 
long,  to  be  used  as  a  pattern  in  making  a  patch-work  pillow-cover.  Con- 
struct the  equilateral  triangle. 

Ex.  44.  Try  to  construct  a  triangle  whose  sides  are  1  in.,  3  in.,  and 
4  in.,  respectively. 

Ex,  45.  Construct  a  triangle  whose  sides  are  2  in.,  3  in.,  and  4  in., 
respectively.  Cut  the  triangle  from  the  paper  and  compare  it  by  super- 
position with  the  triangles  made  by  other  members  of  your  class.  What 
do  you  conclude  must  be  true  about  all  triangles  made  according  to  the 
directions  given  ? 


RECTILINEAR  FIGURES  39 

Proposition  V.     Theorem 

73.  If  two  triangles  have  the  three  sides  of  one  equal 
respectively  to  the  three  sides  of  the  other,  the  triangles 
are  congruent. 


Hypothesis.  In  A  ABC  and  A  DEF : 

AB  =  DE',  BO=EF;  B,ndAC=DF. 

Conclusion.  A  ABC  ^  A  DEF. 

Proof.  1.  Place  A  DEF  so  that  DE  will  coincide  with  AB, 
E  falling  on  B,  and  so  that  F  falls  at  G,  on  the  opposite  side 
of  AB  from  C.     Draw  CG. 

2.  In  A  ACG,  Z 1  =  Z  2,  since  AC=  AG. 

[In  an  isosceles  A,  the  A  opposite  the  equal  sides  are 
equal.]  §  69 

3.  In  A  BCG,  Z  3  =  Z  4,  since  BC=  BG. 

4.  .•.Z1  +  Z3=Z2  +  Z4. 

[If  equals  be  added  to  equals,  the  sums  are  equal.]     Ax.  3  ;  §  51 

5.  .'.ZC=ZG,ovZC  =  ZF.  Ax.7;§51 

6.  In  A  ABC  and  A  DEF: 

AC=  DF,  and  BC=  EF;  Hyp. 

ZC  =  ZF.  Step  5 

7.  .-.  A  ABC  ^  A  DEF. 

[See  Note  1  and  §63.] 

Note  1.  —  From  now  on,  an  authority  which  should  be  familiar  to  the 
student  will  be  omitted  from  demonstrations  in  the  text.  The  paragraph 
reference  will  be  given  for  the  present.  The  student  should  supply  the 
authority  in  full,  without  consulting  the  authority  quoted,  if  possible  ; 
otherwise  he  should  look  up  the  reference.     When  writing  out  a  demon- 


40 


PLANE    GEOMETRY  — BOOK  I 


stration  or  giving  the  demonstration  orally,  give  all  authorities  in  full  as 
has  been  done  in  the  text  heretofore. 

Note  2.  —  Three  sides  determine  a  triangle  ;  that  is,  the  shape  and  size 
of  the  triangle  cannot  change  unless  one  or  more  of  the  sides 
is  changed.      Practical  use  is  made  of  this  fact  as  illustrated 
in  the  figures  below.     In  each  case,  three  lengths  deter- 
mine a  triangle  which  makes  some  part  of  the  object  rigid. 


...^^^^f^. 


Roof  Truss 


P^^ 


-innnnnnnru 
Gate 


lOE^ 


Ex.  46.   If  MN=  iVPand  MO  =  OP,  then  NO  bisects 
/.MNP. 

(Form  the  Hyp.,  Con.,  and  give  the  proof.     See  §  52.) 

Ex.  47.    If    the   opposite    sides   of    a  quadrilateral 
ABGD  are  equal,  then  /.A  =  ZG.  pf ^0 

Ex.  48.  In  quadrilateral  ABCD  in  Ex.  47,  can  you  prove  that  /.B 
=  ZD? 

Ex.  49.  Why  is  a  shelf  bracket  made  in  the  form  of  a  triangle  ? 

Ex.  50.  Can  you  give  any  other  practical  uses  of  Proposition  V  ? 

Ex.  51.   On  segment  XY  construct  isosceles  A  XYZ  and  a  second  isos- 
celes AXYW.      Draw  ZW.      Prove  that  AXZW^ 
A  YZW. 

Ex.  52.    In  the  adjoining  figure,  if  AB  =  DC,   and     b^ 
AC  =  BD,  then  Z  A  must  equal Z  D. 

Suggestion.  —  Prove  A  ABC ^  A  BCD. 
Note.  — Supplementary  Exercises  3  to  5,  p.  273,  can  be  studied  now. 


,^< 


Review  Questions 

Ex.  53.  What  is  a  theorem  ?  an  axiom  ? 

Ex.  54.  State  three  theorems  by  which  two  triangles  can  be  proved 
congruent. 

Ex.  55.  What  are  homologous  parts  of  congruent  triangles  ? 

Ex.  56.  State  Principle  I. 

Ex.  57.  What  does  it  mean  to  bisect  an  angle  ? 


RECTILINEAR    FIGURES  41 

Proposition   VI.     Problem 


74.   Bisect  a  given  angle. 


Given  Z^O-B. 

Required  to  bisect  Z  AOB. 

Construction.  1.  With  0  as  center  and  a  convenient  radius, 
draw  an  arc  intersecting  AO  a,t  C  and  BO  at  D. 

2.  With  C  and  D  as  centers  and  with  equal  radii,  draw  arcs 
intersecting  at  E. 

3.  Draw  OE. 
Statement.                  OE  bisects  Z  AOB. 

Proof.     The  proof  is  to  be  given  by  the  pupil. 

Suggestions.  —  Draw  CE  and  DE.    Recall  §  66. 

Note.  —  For  construction  problems,  the  regular  form  is  to  give  the 
statement  of  the  problem,  the  parts  which  are  "given,"  that  which  is 
"required,"  the  "construction,"  the  "statement,"  and  the  "proof." 
Also  it  is  important  to  "  discuss  "  the  solution  finally,  in  order  to  decide 
when  the  solution  is  possible,  etc.  In  this  problem  it  is  evident  that 
the  solution  is  always  possible. 

Ex.  58.    Draw  an  obtuse  angle.     Divide  it  into  four  equal  parts. 

Ex.  59.  Construct  the  bisectors  of  the  three  angles  of  a  large  triangle. 
What  seems  to  happen  ? 

Ex.  60.  Three  pieces  of  wood  are  to  be  joined  as  in 
the  figure  on  the  right.  Construct  to  scale  (letting  1'' 
=4")  an  equilateral  triangle  ;  bisect  its  angles;  on  each 
bisector  lay  off  a  point  4"  from  the  vertex  ;  connect  these 
points. 


42  PLANE    GEOMETRY  -  BOOK  I 

Proposition  VII.     Problem 

75.    At  a  given  point  in  a  line,  construct  an  angle 
equal  to  a  given  angle. 


^. 


By 


PC  o 

Given  /  P,  and  point  0  in  line  OA. 

Required  to  construct  an  angle  equal  to  Z  P,  having  0  as 
vertex  and  OA  as  side. 

Construction.  1.  With  P  as  center  and  a  convenient  radius, 
draw  an  arc  intersecting  the  sides  of  Z  Pat  (7 and  D,    Draw  CD. 

2.  With  0  as  center  and  PC  as  radius,  draw  arc  FE. 

3.  With  F  as  center  and  CD  as  radius,  draw  an  arc  inter- 
secting arc  FE  at  B. 

4.  Draw  OB. 
Statement.                    Z.AOB  =  ZP. 

Proof.     Draw  FB.     Proof  to  be  given  by  the  pupil. 

Ex.  61.   Construct  /\ABC  with  side  AB  =  4  in.,  and  ZA  and  ZB 
equal  to  the  angles  given  in  the  adjoin- 
ing figure.     Measure  AC.     Should  the 
triangles  made  by  different  pupils  be 
congruent  ? 

Ex  62.  Construct  A  ABC  having  Z  A  equal  to  the  Z  A  given  in  Ex.  61, 
AB  =  3  in.,  and  AC  =  2  in.     Measure  BC. 

Note.  — Supplementary  Exercises  6-9,  p.  273,  can  be  studied  now. 

76.  A  line  perpendicular  to  a  segment  at  its  mid-point  is 
the  Perpendicular-bisector  of  the  segment. 

Ex.  63.  Draw  a  line  BC,  3  in.  in  length.  With  compasses,  locate  a 
point  A  above  BC  which  is  2  in.  from  B  and  2  in.  from  C.  Locate  simi- 
larly a  point  D  below  BC  which  is  3  in.  from  B  and  3  in.  from  C.  Draw 
AD,  cutting  BC  at  E.  (a)  Compare  BE  with  EC  by  means  of  your 
dividers,     (b)  Measure  Z  BE  A.     (c)  What  kind  of  lines  are  AD  and  BC? 


RECTILINEAR  FIGURES 

Proposition  VIII.     Theorem 


43 


77.  If  two  points  are  each  equidistant  from  the  ends 
of  a  segment,  they  determine  the  perpendicular-bisector 
of  the  segment 


Hypothesis.  C  and  D  are  equidistant  from  the  ends  of  seg 
ment  AB.     CD  intersects  AB  at  E. 

Conclusion.  AE  =  EB;     CD±  AB. 

Proof.  1.   In  A  ACD  and  A  CBB : 

AC=CB,kndAD  =  DB', 

CD  =  CD. 
.'.  A  ACD  ^  A  BCD. 
.-.  Z3  =  Z4. 
AACE^AECB. 
(Give  the  full  proof.) 

.-.  AE  =  EB. 

Also     Z1=Z2. 

.-.  CD±AB. 

[If  one  straight  line  meets  another  straight  line  so  that 
the  adj.  A  formed  are  equal,  the  A  are  rt.  A,  and  the  lines 
are  ±.]  §§  20,  29 

Note.  —  It  is  often  necessary,  as  in  this  proof,  to  prove  one  pair  of  tri- 
angles congruent  in  order  to  obtain  two  equal  angles  or  two  equal  segments 
which  are  required  in  turn  to  prove  another  pair  of  triangles  congruent. 

Ex.  64.  If  XZ  is  the  perpendicular-bisector  of  ABj 
and  Y  is  a  point  on  XZ,  prove  A  AXY^  A  BXY. 

First  prove  /\AXZ^/\BXZ  to  get  AX=XB; 
then  prove  A  ^Z  F  ^  A  BZ  T  to  get  ^  F  =  YB. 


2. 
3. 
4. 

5. 

6. 

7. 


Hyp. 

Why? 
§65 
§  63 

Why? 
Why? 


Note.  — Supplementary  Exercises  10-11,  p.  274,  can  be  studied  now. 


44  PLANE   GEOMETRY  — BOOK  I 

.  Proposition  IX.     Problem 
78.   Construct  the  perpendicular-bisector  of  a  given 


segment. 

x1 


V 


a4 


N 


\A 


\ 


■^B 


I  / 

Given  line  segment  AB. 

Required  to  construct  the  perpendicular-bisector  of  AB. 

Construction.     1.   With  A  and  B  as  centers,  and  with  equal 
radii,  draw  arcs  intersecting  at  Q  and  also  at  Z>. 
2.  Draw  CD  intersecting  AB  at  B. 

Statement.  E  bisects  AB. 

Proof.     1.   AC=  BO,  and  also  AD  =  BD. 

[Radii  of  equal  circles  are  equal.]  §  17 

2.  .*.  CD  is  the  perpendicular-bisector  of  AB. 

[If  two  points  are  each  equidistant  from  the  ends  of  a 
segment,  they  determine  the  perpendicular-bisector  of  the 
segment.]  §  77 

Ex.  65.     Divide  a  given  segment  into  four  equal  parts. 
Ex.  66.     Draw  a  triangle  of  large  size.    Construct  the  perpendicular- 
bisectors  of  the  three  sides.     What  happens  ? 

79.  A  Median  of  a  triangle  is  the  line  drawn  from  a  vertex 
to  the  mid-point  of  the  opposite  side. 

Ex.  67.  Draw  a  triangle  of  large  size.  Construct  the  three  medians 
of  the  A.     What  happens  ? 

Ex.  68.  Prove  that  the  median  drawn  to  the  base  of  an  isosceles 
triangle  bisects  the  vertical  angle.     (Construct  the  figure.) 

Note.  —  Supplementary  Exercises  12-14,  p.  274,  can  be  studied  now. 


RECTILINEAR  FIGURES  45 


Proposition  X.     Problem 

SO.  At  a  point  in  a  line,  construct  a  perpendicular 
to  the  line. 


A-J- 


\ 


Given  O,  any  point  in  line  AB. 
Required  to  construct  a  perpendicular  to  AB  at  (7. 
Construction.     1.    With  C  as  center,  and  any  radius,  draw- 
arcs  intersecting  AB  at  D  and  E  respectively. 

2.  With  D  and  E  as  centers  and  a  radius  greater  than  one 
half  DE,  draw  two  arcs  intersecting  at  F. 

3.  Draw  CF. 
Statement.                    CF±  AB  at  C. 
Proof.     1.                    Draw  DF  and  EF. 

2.  C  is  equidistant  from  D  and  E.  Why  ? 

3.  F  is  equidistant  from  D  and  E.  Why  ? 

4.  .-.  CF  is  the  perpendicular-bisector  of  DE.        Why  ? 

5.  .-.  CFJ_^J5at  a 

[Since  AB  and  i>^  are  the  same  straight  line.] 

81.  Proposition  X  proves  that  one  perpendicular  can  be 
draivn  to  a  line  at  a  point  in  the  line. 

It  can  be  proved  that  only  one  perpendicular  can  be  drawn  to 
a  line  at  a  point  in  the  line.  For,  if  CP  and  DP  were  both 
perpendicular  to  AB  at  P,  then  Z  1  and  Z  2  ^  ^ 

would  both  be  right  angles  and  hence  would 
be  equal.  But  Z  1  is  greater  than  Z  2,  for 
the  whole  is  greater  than  any  of  its  parts.    -^ 

Ex.  69.  Prove  CF  perpendicular  to  DE  (§80)  by  proving  that 
Z  FCD  =  Z  FCE,  and  then  using  §  26. 


46  PLANE   GEOMETRY  — BOOK  I 

Proposition  XI.     Problem 

82.   Construct  a  perpendicular  to  a  line  from  a  point 
not  in  the  line, 

G 


•^"i^ 


-fi 


Given  line  AB  and  point  C  not  in  AB. 
Required  to  construct  a  ±  to  AB  from  C. 
Construction.     1.   With  C  as  center  and  a  convenient  radius, 
draw  an  arc  intersecting  AB  at  D  and  E  respectively. 

2.  With  D  and  E  as  centers,  and  equal  radii,  draw  two  arcs 
intersecting  at  F. 

3.  Draw  CF. 

Statement.  CF±AB. 

Proof.     1.  C  is  equidistant  from  D  and  E.  Why  ? 

2.  F  is  equidistant  from  D  and  E.  Why  ? 

3.  .-.  CF  is  the  perpendicular-bisector  of  DE.        Why  ? 

4.  .-.  Ci^±^S. 

[Since  AB  and  J>£'  are  the  same  straight  line.] 

Historical  Note.  —  This  construction  is  attributed  to  Oenipodes  of 
Chios  (466  B.C.) 

83.  Proposition  XI  proves  that  one  perpendicular  can  be 
drawn  to  a  line  from  a  point  not  in  the  line. 

It  will  be  proved  later  (§  88)  that  only  one  perpendicular  can 
he  drawn  to  a  line  from  a  point  not  in  the  line. 

It  will  also  be  proved  (§  164)  that  the  perpendicular  is  the 
shortest  segment  from  the  pfoint  to  the  line. 

Note.  — Supplementary  Exercises  15-16,  p.  274,  can  be  studied  now. 


RECTILINEAR  FIGURES 


47 


84.  The  Distance  from  a  point  to  a  line  is  the  length  of  the 
perpendicular  from  the  point  to  the  line. 

85.  An  Altitude  of  a  triangle  is  the  per- 
pendicular drawn  from  a  vertex  to  the  opposite 
side  or  the  opposite  side  extended ;  as,  AD. 

Ex.  70.  How  many  altitudes  does  a  triangle 
have  ? 

Ex.  71.  Construct  a  triangle  whose  sides  are  2  inches,  3  inches,  and 
4  inches,  respectively.     Construct  the  three  altitudes  of  the  triangle. 

Ex.  72.  Construct  an  angle  of  45°.  (Use  Prop.  XI  and  Prop.  VI.) 
Construct  an  angle  of  135°  ;  of  22^°  ;  of  67^°. 

Note.  —  The  second  and  third  designs  below  are  drawn  upon  the  first 
figure  as  a  background.  Can  you  discover  how  the  first  figure  is  con- 
structed ?    Can  you  make  an  original  design  similar  to  these  designs  ? 


86.  An  Exterior  Angle  of  a  triangle 
is  the  angle  at  any  vertex  formed  by  a 
side  of  the  triangle  and  the  adjacent  side 
extended ;  as,  Z  DC  A. 

One  interior  angle  is  adjacent  to  the  exterior  ^ 
angle  and  the  other  two  are  remote  interior  angles, 
are  the  remote  interior  angles  of  exterior  A  DC  A. 

Ex.  73.   Draw  a  large  figure  like  that  in  §  86.     Measure  the  exterior 
Z.DCA  and  each  of  the  remote  interior  angles.     How  does  the  exterior 

angle  compare  with  the  remote  interior  angles  ? 

• 

Ex.  74.   In  the  figure  for  Prop.  IX,  p.  44,  Z.  CEB  is  an  exterior 
angle  of  what  triangles  ? 


O       D 
Thus,  Z  .A  and  Z.B 


48 


PLANE   GEOMETRY  — BOOK  I 


Proposition^  XII.     Theorem 
87.  A71  exterior  angle  of  a  triangle  is  greater  than 


either  remote  interior  angle. 


Hjrpothesis. 
Conclusion. 


Z  BCD  is  an  exterior  Z  of  A  ABC. 
Z  BCD  >ZB;  also  Z  BCD  >  Z  A. 
Part  I.    Proof.     1.    Through  0,  the  mid-point  of  BC,  draw 
AG.     Extend  AO  to  E,  making  OE  equal  to  AG.     Draw  CE. 


2. 


.-.  AABG^AGCE. 
[Give  the  full  proof.] 

.-.  Z4  =  Z1. 
Z  BCD  >  Zl. 
[The  whole  is  greater  than  any  of  its  parts.] 
.'.ZBCD>Z4.. 
(Substitute  Z4  for  its  equal,  Z  1.) 
Part  IL   Proof.     1.   Extend  BC  to  F. 


Why? 
Ax.  8,  §  51 

Ax.  2,  §  51 


2.  Z  ACF  >  ZA. 

[By  a  proof  similar  to  that  for  Part  I.] 

3.  ZBCD  =  ZACF.  Why? 

4.  .'.  ZBCD>ZA. 

(Substitute  Z 5 (7i>  for  its  equal,  ZAGF,  in  step  2.) 

Ex.  75.    In  the  figure  for  Prop.  XII,  prove  that  ZBOE  is  greater 
than  Z  4.     (Use  §87.) 

Ex.  76.    Prove  also  that  Z  AOC  >  Z  1. 
Ex.  77.    Compare  Z  ECF  with  Z  3. 


RECTILINEAR  FIGURES  '  49 

88.   Cor.     Tliere  can  he  only  one  perpendicular  from  a  point 
to  a  line.  ^ 

If  AB  ±  DE,  then  AC  cannot  be  ±   DE,  for 
Z2  >  A\  and  hence  Z2  is  an  obtuse  angle. 

^-   V         B 

Ex.  78.     If  straight  lines  be  drawn  from  a  point  with-  2 

in  a  triangle  to  the  extremities  of  any  side,  the  angle  in- 
cluded by  them  is  greater  than  the  angle  included  by  the 
other  two  sides.     (Prove  Z  AXC  >ZABC.)  ^'^  ^C 

Suggestions.  —  1.  Compare  Z  AXC  with  Z  A  YC.    2.  Compare  Z  A  YC  with 
ZABC. 


Review  Questions 

Ex.  79.    What  is  the  perpendicular-bisector  of  a  segment  ? 

Ex.  80.  What  is  a  median  of  a  triangle  ?  How  many  medians  does  a 
triangle  have  ? 

Ex.  81.   What  is  an  altitude  of  a  triangle  ? 

Ex.  82.   What  is  the  distance  from  a  point  to  a  line  ? 

Ex.  83.  (a)  What  is  an  exterior  angle  of  a  triangle  ?  (b)  How 
many  exterior  angles  can  be  formed  at  one  vertex  of  a  triangle  ? 
(c)  How  do  they  compare  ? 

Ex.  84.  (a)  How  many  perpendiculars  can  be  drawn  to  a  line  from  a 
point  not  in  the  line  ?     (6)  How  many  at  a  point  of  the  line  ? 

Ex.  85.  Five  fundamental  construction  problems  have  now  been 
taught  (§§  74,  75,  78,  80,  82).  (a)  State  each  of  them.  (6)  Are  you 
able  to  make  each  of  these  constructions  quickly  with  ruler  and  compass 
alone  ? 

Note.  —  Straightedge  and  compass  alone  are  employed  in  making  the 
constructions  in  elementary  geometry.  This  practice  was  initiated  by 
Plato  (429-348  b.c).  Naturally  some  constructions  cannot  be  made 
with  these  tools  alone.  For  example,  it  is  impossible  to  trisect  an  angle 
by  ruler  and  compass  alone. 


50  PLANE   GEOMETRY  —  BOOK  I 

PARALLEL  LINES 


ill  ill  iji  ifi  ifi  fli  -^^ 


Some  Parallel  Line  Border  Designs 

89.  Two  straight  lines  are  Parallel  (II)  if  they  lie  in  the 
same  plane  and  do  not  meet  however  far  they  are  extended. 

Note.  — Two  straight  lines  in  the  same  plane  either  intersect  or  are 
parallel  lines. 

90.  Ax.  16.    Axiom  of  Parallels.    TJirough  a  ^  -^.^p^       ^ 
given  point  there  can  he  only  one  parallel  to  a 
given  line.     Thus,  XFII  MN. 


M N 


91.  Cor.     If  two  lines  are  parallel  to  a     . „ 

third  line,  they  are  parallel  to  each  other. 

Hyp.  ABWGD)  EFWCD.  ^ ^ 

Con.  AB  II  EF. 

Proof.  If  AB  and  EF  are  not  parallel,  they  must  meet  at  a  point. 
Through  this  point  there  would  then  be  two  lines  parallel  to  CD.  But 
this  is  impossible  by  the  Axiom  of  Parallels.  Hence  AB  must  be  paral- 
lel to  EF. 

92.  If  two  lines  are  cut  by  a  third  line,  AB,  called  a  Trans- 
versal, the  angles  are  named  as  follows : 

A  3,  4,  5,  and  6  are  called  Interior  Angles. 

A 1,  2,  7,  and  8  are  called  Exterior  Angles. 

AS    and    6    are    called    Alternate-interior 
Angles ;  also  A  5  and  4. 

Al    and    8    are    called    Alternate-exterior 
Angles ;  also  zi  2  and  7. 

A  2  and  6  are  called  Corresponding  Angles ;  also  /4 1  and  5, 
A  3  and  7,  and  A  4  and  8. 


PARALLEL  LINES 


51 


Proposition  XIII.     Theorem 

93.  If  two  lines  are  cut  hy  a  transversal  so  that  a 
pair  of  alternate-interior  angles  are  equal,  the  lines  are 
parallel. 


>o 


Hypothesis.     AB  and  CD  are  cut  by  EF-,  Z  1  =  Z  2. 

Conclusion.  AB  II  CD. 

Proof.  1.  Suppose  that  AB  is  not  parallel  to  CD,  and  that 
it  meets  CD  at  point  O  on  the  right  of  EF,  forming  A  MNO. 

2.  Then  Z  1  is  an  exterior  angle  of  A  MNO. 

3.  .-.  Z  1  >  Z  2 

[An  ext.  Z  of  a  A  is  greater  than  either  remote  int.  Z.]         §  87 

4.  ButZl  =  Z2  Hyp. 

5.  .*.  AB  cannot  meet  CD  on  the  right  of  EF. 

6.  Similarly  it  can  be  proved  that  AB  cannot  meet  CD  on 
the  left  of  EF. 

7.  .-.  AB  II  CD 

[Two  lines  are  II  if  they  lie  in  the  same  plane  and  do  not 
meet  however  far  they  are  extended.]  §  89 

Note.  —  Read  the  next  paragraph  at  this  time. 

94.  The  method  of  proof  used  in  §§91  and  93  is  called  the 
Indirect  Method  of  Proof.  Notice  :  (a)  it  starts  by  assuming 
the  negative  of  the  conclusion  ;  (b)  it  follows  up  the  conse- 
quences of  this  assumption  until  a  statement  is  reached  which 
contradicts  a  known  fact  ;  (c)  this  contradiction  is  made  the 
basis  for  asserting  that  the  desired  conclusion  is  true. 


52 


PLANE    GEOMETRY  — BOOK  I 


95.  Principle  II.     To  prove  two  lines  parallel,  try  to  prove 
a  pair  of  alternate-interior  angles  equal. 

96.  Cor.  1.  If  two  lines  are  cut  by  a 
transversal  so  that  a  pair  of  corresponding 
angles  are  equal,  the  lines  are  parallel. 

Hyp.     AB  and  CD  are  cut  by  EF\  Z2=  Z6. 

Con.  AB  li  CD. 

Plan.     Try  to  prove  Z  3  =  Z  6.    Then  use  §  93. 

97.  Cor.  2.     If  two  lines  are  perpendic- 
ular to  a  third  line,  they  are  parallel. 


Hyp.     AB±XY;   CD  A.  XY. 

Con.  AB  II  CD. 

Plan.    Try  to  prove  Z  1  =  Z2.   Then  use  §  93. 


X— f- 


-M- 


B 


D 


98.  Cor.  3.  If  two  lines  are  cut  by  a  transversal  so  that  a 
pair  of  interior  angles  on  the  same  side  of  the  transversal  are 
supplementary,  the  lines  are  parallel. 

Hyp.   AB  and  CD  are  cut  by  EF ;  Z  4  +  Z  6  =  1  st.  Z. 
Con.  AB  II  CD. 

Plan.    Try  to  prove  Z  3  =  Z  6. 
Proof.    1.       Z3  +  Z4  =  1  St.  Z. 

2.  Z4  +  Z6  =  1  St.  Z. 

3.  .-.  Z3  +  Z4  =  Z4  +  Z6. 

4.  .-.  Z3  =  Z6,  and  hence^^ll  CD. 


§39 

A 

Hyp. 

C 

Why? 

Why? 

Ex.  86.  If  tw^o  lines  are  cut  by  a  transversal  so  that  a  pair 
of  alternate-exterior  angles  are  equal,  the  lines  are  parallel. 

Ex.  87.  A  carpenter  wants  a  board  of  length  AC  v^ith 
parallel  ends,  AB  and  CD.  He  marks  AB  and  CD  by  means  of 
his  square  as  in  the  figure.  Why  must  AB  and  CD  be  parallel  ? 
Assume  that  edge  ^O  is  a  straightedge. 


TZyp 


A  B 


CD 


Ex.  88.  If  sides  BA  and  CA  of  any  A  ABC  are  extended  their  own 
lengths  through  vertex  ^  to  Z>  and  E  respectively,  then  DE  is  parallel  to 
BC. 

Suggestion.  —  Apply  §  95  and  §  66. 

Note.  —  Supplementary  Exercises  17-21,  p.  274,  can  be  studied  now. 


PARALLEL  LINES 


53 


Proposition  XIV.     Problem 

99.    Construct  a  parallel  to  a  line  through  a  point 
not  in  the  line.  e 


Given  line  AB  and  (7,  any  point  not  in  line  AB. 
Required  to  construct  a  parallel  io  AB  through  C. 
Construction  (a).     1.   Draw  FE  through  (7,  meeting  AB  at  O. 
2.  At    (7,  construct  Z  2=  Z  1. 

Statement.  MN II  AB.  Why  ? 

Proof.     To  be  given  by  the  pupil. 

Construction  (6).     A  second  construction  is  based  upon  §  96. 
This  construction  is  left  as  an  exercise  for  the  pupil. 


Ex.  89.  The  figure  adjoining  shows 
how  a  draughtsman  draws  a  line  through 
C  parallel  to  AB.  Why  is  DE  parallel  to 
ABf 


Ex.  90.  The  adjoining  figure  shows 
how  a  draughtsman  draws  parallel  lines 
by  means  of  his  T-square.  Why  are  the 
lines  parallel  ? 


Ex.  91.  Construct  the  figure  for  §  07  ;  then  construct  the  bisectors  of 

A  1  and  2.  Prove  that  these  bisectors  are  parallel. 

Ex.  92.  In  the  figure  for  Prop.  XII,  p.  48,  prove  that  CE  \a  parallel 
to  AB. 


54 


PLANE   GEOMETRY  — BOOK  I 

Proposition  XY.     Theorem 


100.   If  tivo  parallels  are  cut  hy  a  transversal,  alter- 
nate-interior angles  are  equal. 


Hypothesis.     EFcuts  lis  AB  and  CD  at  G  and  H. 

Conclusion.     Z  AGH  =  Z  GHD. 

Proof.     1.    Suppose  that  Z  AGH  is  less  than  Z  GHD. 

2.  Then  draw  i^l^  through  H,  so  that  Z  (7^3f  =  Z  ^(^^. 

3.  .'.LMWAB. 

[If  two  lines  are  cut  by  a  transversal  so  that  a  pair  of  alt.- 
int.  A  are  equal,  the  lines  are  II.] 

4.  But  this  is  impossible,  for  CD  II  AB,  by  hypothesis. 

[Through  a  given  point,  there  can  be  only  one  II  to  a 
given  line.] 

5.  .-.  Z  AGH  cannot  be  less  than  Z  GHD. 

6.  Similarly,  Z  AGH  cannot  be  greater  than  Z  GHD. 

7.  .'.  Z  AGH  =  Z  GHD. 
Note.  —  Review  §  94. 


93 


90 


E 


Ex.  93.     If  EF  cuts  parallels  AB  and  CD  so    ^ 
that  ZS  =  30°,  how  many  degrees  are  there  in  each 
of  the  other  angles  of  the  figure  ?  G 


Ex.  94.  If  EF,  joining  two  parallels,  be  bisected 
and  GH  be  drawn  through  the  mid-point  and  included 
between  the  parallels,  then  GH  will  also  be  bisected 
by  the  point. 


1/2 

j> 

3/4 
5/6 

j\ 

7/8 

F 

G 

E 

K 

PARALLEL  LINES 


55 


101.     Cor.  1.   If  two  parallels  are  cut  by  a  transversal^  cor- 
responding angles  are  equal. 


Hyp.     AB  II  CD;  Z2  and  Z.Q  are  corresponding 
angles. 

Con.    Z  2  =  Z  6. 

[What  do  you  know  about  Z  3  and  Z  6  ?] 


102.  Cor.  2.  If  a  line  is  perpendicular 
to  one  of  two  parallels,  it  is  perpendicular  to 
the  other  also. 

Hyp.     AB  II  CD  ;  XY  ±  AB. 
Con.  XT±CD. 

[What  must  be  proved  about  Z  2  ?] 


103.   Cor.  3.   If  two  parallels  are  cut  by  a  transversal,  intenor 
angles  on  the  same  side  of  the  transversal  are  supplementary. 

Hyp.    AB  II  CD ;  Z  4  and  Z  6  are  int.  A  on  the  same  side  of  the 
transversal.     (Fig.  §  101.) 

Con.  Z  4  +  Z  6  =  1  St.  Z. 

Proof.     1.  Z 4  +  Z3  =  1  St.  Z.  §  39 

2.  Z6  =  Z3.  Why? 

8.  .-.  Z4  +  Z6  =  l  St.  Z. 

[Substituting  for  Z  3  its  equal,  Z  6.] 


Ax.  2,  §  51 

\ Isi  Ave. 

'  ^%\      2nd  Ave, 


Ex.  95.  If  N.  W.  Street  crosses  1st  Ave.  at  an 
angle  of  45°,  at  what  angle  does  it  cross  the  parallel 
avenues  ? 


Ex.  96.     In  the  adjoining  figure,  if  AB  II  CD^ 
and  EFWGH,  prove:   (a)  Z  1  =  Z  13  ; 

(6)  Z  3  +  Z  16  =  1  St.  Z.  "     11/12  15/16 

F       H 

Ex.  97.  If  a  line  be  drawn  parallel  to  the  base  of  an  isosceles  triangle, 
cutting  the  two  sides  of  the  triangle,  it  makes  equal  angles  with  these 
sides. 


56 


PLANE   GEOMETRY  — BOOK  I 


104.  One  theorem  is  called  the  Converse  of  another  when  the 
hypothesis  and  conclusion  of  the  one  become  the  conclusion  and 
hypothesis  of  the  other.  Thus,  Prop.  XV  is  the  converse  of  • 
Prop.  XIII. 


In  Proposition  XIII 

In 

Proposition  XV 

Hyp.  EFcMtsABsiud 

Hyp. 

FF  cuts  AB  and 

A 

CD. 

CD. 

Z3  =  Z6. 

AB  11  CD. 

C 

Con.    ABWCD. 

Con. 

ZS=Z6. 

The  converse  of  a  given  theorem  is  not  always  true. 
Thus,  all  right  angles  are  equal.     The  converse  would  be  all 
equal  angles  are  right  angles.     Evidently  this  is  not  true. 


Ex.  98.  Of  what  statement  is  Cor.  1  (§  101)  the  converse  ?  Cor.  2 
(§  ]02)  ?     Cor.  3  (§  103)  ? 

Ex.  99.   In  the  figure  for  §  101,  prove  Z  3  =  Z  7. 

Ex.  100.    In  the  figure  for  §  101,  prove  Z  3  +  Z  5  =  1st  Z. 

Ex.  101.  If  two  parallels  are  cut  by  a  transversal,  alternate  exterior 
angles  are  equal. 

Ex.  102.    State  the  converse  of  Ex.  101.    Is  it  a  true  statement  ? 

Ex.  103.  If  two  parallels  are  cut  by  a  transversal,  bisectors  of  a  pair 
of  corresponding  angles  are  parallel. 

Ex.  104.  If  two  parallels  are  cut  by  a  transversal,  exterior  angles  on 
the  same  side  of  the  transversal  are  supplementary. 

Suggestion.  —  The  proof  is  like  that  for  §  103. 

Ex.  105.   State  the  converse  of  Ex.  104. 

Ex.  106.  If  two  lines  are  perpendicular  to  parallel  lines,  then  they 
are  either  parallel  or  coincide. 

Ex.  107.   What  is  the  axiom  of  parallels  ? 

Ex.  108.   What  are  parallel  lines  ? 

Ex.  109.  Is  the  following  a  correct  statement  ?  When  two  lines  are 
cut  by  a  transversal,  alternate  interior  angles  are  equal. 

Note.  —  The  various  theorems  about  angles  made  by  a  transversal  of 
two  parallels  are  found  in  Euclid.  They  were  probably  formulated  by 
the  Pythagoreans. 


PARALLEL  LINES  57 

Proposition  XVI.     Theorem 
105.   If  tioo  angles  have  their  sides  i^espectively  par- 
allel, they  are  equal,  provided  both  pairs  of  parallels 
extend  in  the  same  directions  from  their  vertices,  or  in 
opposite  directions. 


/ 


'-7 


E 


Fig.  1  Fig.  2 

I.  (Fig.  1.)  Hypothesis.  A  ABC  and  DEF  have  AB  II  DE 
and  BC  II  EF. 

Conclusion.  Z  B  = /.  E. 

[Proof  to  be  given  by  the  pupil.] 

Suggestion.  —  Extend  BC  and  ED  until  they  intersect  at  G.  Compare  Z.  B 
with  Z  2  and  Z  E  with  Z  2.    Then  compare  Z  B  with  Z  £". 

II.  (Fig.  2.)  Hypothesis.  ^  ABC  and  i>^i^  have  ^B  II  DE 
and  5C  II  EF: 

Conclusion.  Z  B  =  ZE. 

Note.  —  The  sides  extend  in  the  same  direction  if  they  are  on  the  same 
side  of  a  straight  line  joining  their  vertices,  and  in  opposite  directions  if 
they  are  on  opposite  sides  of  this  line. 

Ex.  110.   If  two  angles  have  their  sides  respectively 
parallel,  one  pair  of  parallels  extending  in  the  same 
directions  but  the  other  pair  extending  in   opposite 
directions  from  their  vertices,  the  angles  are  supple-     f-*- 
mentary.     (Prove  Z  B  -\-  Z  E  =  I  at.  Z.) 

Ex.  111.  Two  streets  cross  as  in  the  adjoining 
figure.  If  the  lot  lines  at  corner  C  make  an  angle  of 
70°,  determine  the  number  of  degrees  in  the  angle 
formed  at  each  of  the  other  corners. 


J^ 


58 


PLANE   GEOMETRY  — BOOK  I 


Proposition  XVII.     Theorem 

106.    The  sum  of  the  angles  of  any  triangle  is  one 
straight  angle. 


Hypothesis.  A  ABC  is  any  triangle. 

Conclusion.  ZA-\-ZB-\-ZO=lst  Z. 

Proof.    1.   Extend  AC  to  D,  and  construct  CE  parallel  to  AB. 

2.  Z  1  +  Z  2  +  Z  3  =  1  St.  Z. 

[The  sum  of  all  the  successive  adj.  A  around  a  point 
on  one  side  of  a  st.  line  is  one  st.  Z.]  §  34 

3.  Z  ^  =  Z  1.  Const. 

4.  Z  B  =  Z2,  since  BC  cuts  lis  AB  and  CE.  Why  ? 

5.  .'.ZA  +  ZB  +  ZC=lst.Z. 
[Substituting  ZAtor  Zl.ZB  for  Z 2,  and  Z  C for  Z 3.]     Ax. 2,  § 51 

Note.  —  This  theorem  is  attributed  to  Eudemus,  a  pupil  of  Aristotle. 

Ex.  112.     If  Z  ^  =  70°,  and  Z  5  =  35°,  how  large  is  Z  C  ? 

Ex.  113.     How  large  is  each  angle  of  an  equiangular  triangle  ? 

Ex.  114.    Prove  Prop.  XVII  by  constructing  a  line  through  B  parallel 
to^C 

Ex.  115.  The  rafters  of  a  "saddle  roof  "  make 
an  angle  of  40°  with  a  level  line.  What  angle  do 
the  rafters  form  at  the  ridge  ? 


Ex.  116.     AB 

so  that  AD  =  AB. 


AC  in  A  ABC.     BA  is  extended  to  D 
Prove  that  CD  is  perpendicular  to  BC. 


Suggestions.  —  1.   Z 1  +  Z  4  must  be  proved  a  right  Z. 
2.   Whatpartof  Zl  +  Z2  +  Z3  +  Z4isZH-Z4? 

Note.  — This  is  a  very  important  exercise.  It  may  be  expressed  thus: 
if  the  median  to  one  side  of  a  triangle  is  one  half  of  that  side,  the  angle 
from  which  it  is  drawn  is  a  right  angle. 


PARALLEL  LINES  59 

107.  A  triangle  is  a  Right  Triangle  when  it  has  one  right 
angle. 

The  Hypotenuse  of  a  right  triangle  is  the  side  opposite  the 
right  angle ;  the  Legs  of  a  right  triangle  are  the  two  sides 
of  the  triangle  including  the  right  angle. 

If  the  legs  of  a  right  triangle  are  equal,  the  triangle  is 
called  an  Isosceles  Right  Triangle. 

Corollaries  to  Proposition  XVII 

108.  Cor.  1.  A  triangle  cannot  have  two  right  angles  or  two 
obtuse  angles. 

109.  Cor.  2.  The  acute  angles  of  a  right  triangle  are  com- 
plementary. 

110.  Cor.  3.     An  exterior  angle  of  a  triangle     /-^^ 
equals  the  sum  of  the  two  remote  interior  angles.        14 

Prove  Zl  =  Z3  +  ^4. 

[Z2  +  Z1  =  ?    Z2  +  Z3  +  Z4  =  ?    Form  an  equation.] 

111.  Cor.  4.  If  two  angles  of  one  tri- 
angle equal  respectively  two  angles  of 
another  triangle,  the  third  angles  are 
equal. 

Hyp.  Z  1  =  Z  4,  and  Z  2  =  Z  5. 

Con.  Z3  =  Z6. 

112.  Cor.  5.     If  two  triangles  have  a  side,  the  opposite  angle, 
and  another  angle  of  the  one  equal  re-  „ 
spectively  to  a  side,  the  opposite  angle, 

and  another  angle  of  the  other,  the  tri-      

angles  are  congruent. 

Hyp.  AB  =  DE\  Z.C  =  ^F;  ZB  =  ZE. 

Con.  AABC^ADEF. 

Suggestions.  —  1.  Prove  LA  =  LB. 

2.   Then  prove  /S  ABC^/^  DBF  by  §  63. 

Note.  —  Supplementary  Exercises  22-35,  p.  275,  can  be  studied  now. 


60 


PLANE    GEOMETRY  —  BOOK  I 


Propositiot^  XYIII.     Theorem 

113.    If  two  angles  have  their  sides  respectively  per- 
jpendicular,  they  are  either  equal  or  supplementary. 


v"   \ 

JB 


G 
Hypothesis.  A  ABC  and  EDF  have 

ABJuDE  and  BC±FG. 

Conclusion.  Z7  =  Z.1. 

Z  7  +  Z  2  =  1  St.  Z. 

Proof.     1.   Extend  DE  until  it  meets  BA  at  B. 
until  it  meets  EG  at  S  and  intersects  DB  at  0. 

2.  In  A  BBO  said  A  ODS : 

Z6  =  Z5; 
Z  3  =  Z  4. 

3.  .-.ZT^Zl. 

4.  Z  1  -h  Z  2  =  1  St.  Z. 

5.  .-.  Z7  +  Z2==  1  St.  Z. 
[Substitute  Z  7  for  its  equal  Z  1.] 


Extend  BO 


Why? 

Why? 

§111 

Why? 

Ax.  2,  §  51 
They 


Note.  —  The  angles  are  equal  if  both  are  acute  or  both  obtuse 
are  supplementary  if  one  is  acute  and  one  is  obtuse. 

Ex.  117.  If  two  right  triangles  have  the  hypotenuse  and  an  acute 
angle  of  one  equal  respectively  to  the  hypotenuse  and  an  acute  angle  of 
the  other,  they  are  congruent.     (§  112.) 

Ex.  118.  If  two  right  triangles  have  a  leg  and  the  opposite  acute  angle 
of  one  equal  respectively  to  a  leg  and  the  opposite  acute  angle  of  the  other, 
they  are  congruent. 


PARALLEL  LINES 


61 


Proposition  XIX.     Theorem 

.  114.  If  two  right  triangles  have  the  hypotenuse  and 
a  leg  of  one  equal  respectively  to  the  hypotenuse  and  a 
leg  of  the  other ,  the  triangles  are  congruent. 


Hypothesis.     In  rt.  A  ABC  and  DBF-. 

hypotenuse  AB  =  hypotenuse  DE ;  BC=  EF. 
Conclusion.  AABC^ADEF. 

Proof.  1.  Place  A  ABjO  beside  A  DEF  so  that  BC  will 
coincide  with  its  equal  EF,  B  falling  on  E,  and  so  that  A 
falls  at  Gj  on  the  opposite  side  of  EF  from  D. 

2.  .-.  Zl +  Z2  =  1  St.  Z.  Why? 

3.  .-.  DFG  is  a  straight  line.  §  40 

4.  .-.  figure  EDFG  is  a  triangle. 

.6.  .'.  ^G  =  ZD,  ov  Z.A=Z.D.  §69 

6.  In  A  ABC  and  A  DEF: 

AB  =  DE:  Z1  =  Z2:  ZA=ZD. 


A  ABC  ^  A  DEF. 


§112 


115.   Cor.     If  two  equal  oblique  segments  are  di'awn  to  a  line 
from  a  point  in  a  perpendicular  to  the  line : 

(1)  they  cut  off  equal  distances  from  the 
foot  of  the  perpendicular.    (Prove  AD=DB.) 

(2)  they  make  eqtial  angles  with  the  per- 
pendicular.    (Prove  Z  1  =  Z  2.) 

(3)  they  make  equal  angles  with  the  given 
line.     (Prove  Z  3  =  Z  4.) 


Note.  —  Supplementary  Exercise  36,  p.  276,  can  be  studied  now. 


62  PLANE-  GEOMETRY  —  BOOK   I 

SUMMARY 

116.   The  student  will  liave  constant  use  for  the  foregoing 
theorems,  problems,  and  facts. 

A.  Two  triangles  are  congruent  if  : 

1.  Two  sides  and  the  included  Z.  of  one  are  equal  respectively,  etc.  §  63 

2.  Two  A  and  the  included  side  of  one  are  equal  respectively,  etc.  §  67 

3.  The  three  sides  of  one  are  equal  respectively,  etc.  §  73 

4.  A  side,  the  opposite  Z,  and  another  Z  of  one  are  equal  respectively, 
etc.  §  112 

B.  Two  right  triangles  are  congruent  if  : 

1.  The  hypotenuse  and  a  leg  of  one  are  equal  respectively,  etc.     §  114 

2.  The  hypotenuse  and  an  acute  A  of  one  are  equal  respectively,  etc. 

Ex.  117 

3.  A  leg  and  the  opposite  acute  Z  of  one  are  equal  respectively,  etc. 

Ex.  118 

C.  Two  lines  are  parallel  if  : 

1.  Alt. -int.  A  made  by  a  transversal  are  equal.  §  93 

2.  Corresponding  A  made  by  a  transversal  are  equal.  §  96 

3.  Int.  A  on  the  same  side  of  the  transversal  are  supp.  §  98 

4.  They  are  parallel  to,  or  perpendicular  to,  the  same  line.  §§  91,  97 

D.  If  a  transversal  cuts  two  parallels  : 

1.  Alt. -int.  A  are  equal.  §  100 

2.  Corresponding  A  are  equal.  §  101 

3.  Int.  A  on  the  same  side  of  the  transversal  are  supp.  §  103 

E.  To  prove  two  line  segments  are  equal : 

Try  to  prove  them  homologous  sides  of  congruent  A.  §  66 

E.  To  prove  two  angles  equal,  try  to  prove  that  they  : 

1.  Are  homologous  A  of  congruent  A.  §  66 

2.  Are  supplements  or  complements  of  the  same  or  equal  A.     §§  37,  41 

3.  Are  right  A  or  vertical  A.  '         §§  27,  54 

4.  Are  opposite  the  equal  sides  of  an  isosceles  A.  §  69 

5.  Are  alt. -int.  A  or  corresponding  A  made  by  a  transversal  of  two 
parallels.  §§  100,  101 

6.  Have  their  sides  respectively  II  or  ±,  etc.  §§  105,  113 
G.   To  prove  an  angle  is  a  right  angle,  try  to  prove  : 

1.  It  is  equal  to  its  supplement.  §  26 

2.  It  is  the  angle  formed  by  two  lines  which  are  -L  by  §  77. 

3.  It  is  equal  to  an  angle  known  to  be  a  right  Z. 


SUMMARY  63 

117.  Success  in  demonstrating  unproved  theorems  comes  as 
a  result  of  knowledge  of  the  facts  summarized  in  §  116,  system- 
atic methods  of  studying  a  theorem  and  planning  its  demon- 
stration, and  experience  and  perseverance. 

Directions 

1.  Read  the  theorem  carefully,  making  certain  that  each 
word  is  thoroughly  understood. 

2.  Draw  the  figure  carefully,  constructing  it  when  possible. 

Make  the  figure  general.  Thus,  if  the  figure  is  based  upon  a  triangle, 
do  not  draw  a  right  triangle  or  an  isosceles  triangle  unless  told  to  do  so. 

3.  Decide  upon  the  hypothesis  and  conclusion. 

(a)  Remember  that  the  hypothesis  states  the  facts  about  the  figure 
which  are  assumed,  and  that  the  conclusion  states  the  facts  which  are  to 
be  proved. 

(6)  If  the  theorem  is  stated  in  the  "  if  .  .  .  then  .  .  ."  form  (§  52), 
the  hypothesis  and  conclusion  are  evident  at  once. 

(c)  If  the  theorem  is  not  stated  in  the  "  if  .  .  .  then  ..."  form,  the 
declarative  sentence  in  its  simplest  form  will  give  the  conclusion,  and  the 
subject  of  the  sentence  with  its  modifiers  will  give  the  hypothesis. 

4.  Decide  upon  a  plan  for  the  demonstration. 

For  the  present,  the  suggestions  in  §  116  will  aid  the  pupil 
in  solving  most  exercises. 

Ask  "  what  does  the  conclusion  mean  ?  "  or  "  how  can  I  prove 
the  conclusion  ? "  The  answers  will  suggest  a  plan  for  the 
proof. 

Thus,  suppose  that  the  conclusion  is  :  EA  =  EB. 
Question.     How  can  I  prove  EA  =  EB  ? 
Answer.     By  proving  them  homologous  parts  of  congruent  ^. 

This  means  that  two  triangles  of  which  EA  and  EB  are  sides  must  be 
selected. 

Question.     How  can  I  prove  two  triangles  congruent? 
Answer.     By  one  of  the  methods  given  in  §  116,  A  and  B. 

This  leads  to  the  comparison  of  the  sides  and  angles  of  the  triangles. 
Question.     What  do  I  know  about  the  sides  and  A  of  the  ^  ?     How  can 

I  prove  these  two  angles  equal  ? 
Answer.    §  116,  E  and  F  suggest  possible  answers. 


64 


PLANE    GEOMETRY  —  BOOK   I 


Proposition^  XX.     Theorem 

118.  I.    Any  point  in  the  p)erpendicular-bisector  of  a 
segment    is    equidistant  from    the 

ends  of  the  segment. 

Hypothesis.     CD  1.  AB -,   AD  =  DB ; 

E  is  any  point  in  CD. 

Conclusion.     EA  —  EB. 

Plan.  Try  to  prove  EA  and  ^5  horn, 
sides  of  cong.  A. 

[Proof  to  be  given  by  the  pupil.] 

IT.  (Converse.)  Any pohit  equi- 
distant from  the  ends  of  a  segment 
lies  in  the  perpeyidicular-hisector  of 
the  segment. 

Hypothesis.      AB  is  a  st.  line. 
PA  =  PB. 

Conclusion.     F  lies  in  the  perpendicular-bisector  of  AB. 

Plan.     Let  C   be   the   mid-point    of  AB.      Try    to    prove 
PC  A.  AB,  by  proving  Z  1  =  Z  2. 
Proof.     1.  AAPC^APCB. 

[Give  the  full  proof.] 

2.  .-.  Z1  =  Z2.  Why? 

3.  .:PC±AB. 

[If  one  St.  line  meets  another  st.  line  so  that  the  adj.  A 
formed  are  equal,  the  A  are  rt.  A  and  the  lines  are  ±.] 

§§  26,  29 

119.  Cor.  Two  obliques,  drawn  to  a  line  from  a  point  in  a 
perpendicular  to  the  line  and  cutting  off  equal  distances  from  the 
foot  of  the  perpendicular,  are  equal. 

Note.  —  Supplementary  Exercises  37-38,  p.  276,  can  be  studied  now. 


TRIANGLES 


65 


120. 


Proposition  XXI.     Theorem 
I.   Any  point  in  the  bisector  of  an  angle  is  equi- 


distant from  the  sides  of  the 
angle. 

Hypothesis.  BD  bisects 
ZABC;  Pisin  BD'yPM±AB; 
PjSr±Ba 

Conclusion.     PM  =  PN. 

Plan.     Try  to  prove  PM  and  P^hom.  parts  of  cong.  A. 

Suggestion.  —  Recall  §  112. 

II.    (Converse.)     A?i2/  point  equidistant  from   the 
sides  of  an  angle  lies  in  the  bi- 
sector of  the  angle. 

Hypothesis.  P lies  within  Z  ABC\ 
PM1.AB',  PN±BC;  PM=PN. 
Conclusion.     P  lies  in  the  bisector 

of  Z  ABC. 
Plan.    Draw  PB.     Try    to  prove 
PB  bisects  Z  ABC^  by  proving  Z  3  =  Z  4. 

121.  Cor.  Any  point  not  in  the  bisector  of  an  angle  is 
unequally  distant  from  the  sides  of  the  angle. 

Note.  — Supplementary  Exercise  39,  p.  276,  can  be  studied  now.  • 

ISOSCELES  AND  EQUILATERAL  TRIANGLES 

122.  Review  the  definitions  of  isosceles  and  equilateral 
triangles  in  §  68;   also  review  Prop.  Ill,  §  70,  and  Ex.  113. 

Notice  that  an  equilateral  triangle  is  a  special  form  of 
isosceles  triangle.  Hence,  for  each  theorem  about  an  isosceles 
triangle  there  is  a  corresponding  theorem  about  an  equilateral 
triangle,  which  may  be  considered  a  corollary  of  the  former. 
Thus,  §  70  follows  at  once  from  Prop.  III. 


66 


PLANE    GEOMETRY  —  BOOK   I 


Proposition  XXII.     Theorem 

123.  If  two  angles  of  a  triangle  are  equal,  the  sides 
opposite  are  equal,  and  the  triangle  is  isosceles. 


Hypothesis.  In  A  ABC,  ZA  =  ZB. 

Conclusion.  AC  =  EC. 

Plan.     Try  to  prove  AG  and  BC  horn,  sides  of  cong.  A. 
Proof.     1.    Construct  CD  bisecting  Z  ACB, 

[Complete  the  proof  in  good  form.  ]  • 

124.   Cor.     If  a  triangle  is  equiangular,  it  is  also  equilateral. 

Ex.  119.     Prove  that  the  bisector  of  the  vertical  angle  of  an  isosceles 
triangle  is  perpendicular  to  and  bisects  the  base. 

Ex.  120.     Prove  that  the  altitude  to  the  base  of  an  isosceles  triangle  is 
also  the  median  to  the  base  and  bisects  the  vertical  angle. 

Ex.  121.     Prove  that  the  altitudes  drawn  to  the  eqaal  sides  of  an 
isosceles  triangle  are  equal. 

Ex.  122.     Prove  that  the  medians  drawn  to  the  equal  sides  of  an 
isosceles  triangle  are  equal. 

Ex.  123.     A  boy  wishes  to  make  a  saw-buck.     Assume  that  B0=  OD 
and  that  Z  EPF  =  40°.     Determine  the  angles  at  B 
and  D  so  that  the  pieces  AB  and  CD  will  stand  firmly 
upon  the  ground.     Determine  the  angles  at  C  and  A  so 
that  CE  and  FA  will  be  parallel  to  the  ground  line. 

Ex.  124.     Construct  an  angle  of  60°. 

(Construct  an  equilateral  triangle.) 
Also  construct  an  angle  of  :  30°  ;  120°  ;  150°. 

Ex.  125.     If  one  angle  of  an  isosceles  triangle  is  60°,  the  triangle  is 
equilateral. 


TRIANGLES 


67 


Ex.  126.  The  bisectors  of  the  equal  angles  of  an 
isosceles  triangle  form  with  the  base  another  isosceles 
triangle. 

Ex.  127.     If  the  bisector  of  the  exterior  angle  at  one      ^^ ^B 

vertex  of  a  triangle  is  parallel  to  the  side  joining  the  other  two  vertices, 
the  triangle  is  isosceles. 

Ex.  128.  If  one  acute  angle  of  a  right  triangle  is 
30°,  the  side  opposite  is  one-half  the  hypotenuse. 

Suggestion.  —  Extend  BC  to  Z),  making  CD  equal  to 

BC.    Prove  A  ABD  is  equilateral.  _ 

B 

Note. — This  is  a  very  important  exercise.     Pupils 
should  endeavor  to  remember  it,  as  it  will  be  required  in  the  proofs  of 
certain  exercises  in  geometry. 

Ex.  129.     Prove  that  the  perpendiculars  drawn  from  the  mid-point  of 
the  base  of  an  isosceles  triangle  to  the  sides  of  the  triangle  are  equal. 

Ex.  130.  If  the  equal  sides  of  an  isosceles  triangle 
be  extended  beyond  the  base,  the  exterior  angles  so 
formed  are  equal. 


Ex.  131.  Prove  that  the  bisector  of  the  exterior  angle 
at  the  vertex  of  an  isosceles  triangle  is  parallel  to  the 
base  of  the  triangle. 

Suggestion.  — Comi^2ire  LBCD  with  LA-^  LB  (§  110). 


Ex.  132.     In  the  gable  in  the  front  of  a  garage, 
the  two  boards  whose  upper  edges  are  AB  and  A  C 
are  of  equal  length  and  meet  at  a  point  .<4  on  a  line  'b 
AD  which  is  perpendicular  to  BC. 

If  ZACD  =  30°,  how  large  are  Z  ABD,  Z  CAD, 
and  Z  BAD  ? 


Ex.  133.     If  the  perpendiculars  drawn  from  the  mid-point  of  one  side 
of  a  triangle  to  the  other  two  sides  are  equ^,  the  triangle  is  isosceles. 

Ex.  134.     If  the  altitudes  drawn  to  two  sides  of  a  triangle  are  equal, 
the  triangle  is  isosceles. 


r- 

/ 

E 

\ 

.1 

/ 

\. 

Note.  — Supplementary  Exercises  40-49,  p.  276,  can  be  studied  now, 


68 


PLANE    GEOMETRY  —  BOOK  I 


POLYGONS 

125.  A  Polygon  is  a  dosed  (§  6)  broken  line ;   as  ABCDE, 
Points  A,  B,  C,  etc.,  are  the  vertices  of 

the  polygon ;  A  A,  B,  C,  etc.,  are  the 
angles  ;  AB,  BC,  CD,  etc.,  are  the  sides  ; 
the  sum  of  the  lengths  of  the  sides  is  the 
perimeter  oi  the  polygon;  a  line  joining 
any  two  non-consecutive  vertices  is  a  di- 
agonal of  the  polygon ;  as  ^C. 

A  polygon  incloses  a  portion  of  the  plane  called  the  in- 
terior of  the  polygon. 

126.  A  polygon  is  Convex  if  no  side, 
when  extended,  will  pass  through  the  inte- 
rior of  the  polygon  ;  as  ABCDE  of  §  125. 

A  polygon  is  Concave  if  at  least  two 
sides,  when  extended,  will  pass  through  the 
interior  of  the  polygon ;  as  FGHIK. 

127.  Only  convex  polygons  are  considered  in  this  text.  A 
convex  polygon  having  n  sides  has  n  vertices. 

128.  An  Equilateral  Polygon  is  one  whose  sides  are  all  equal. 
An  Equiangular  Polygon  is  one  whose  angles  are  all  equal. 

129.  Two  polygons  are  mutually  equilateral  if  the  sides  of 
one  are  equal  respectively  to  the  sides  of  the  other  ;  and  mutu- 
ally equiangular  if  the  angles  of  one  are  equal  respectively  to 
the  angles  of  the  other.  If  two  polygons  are  both  mutually 
equiangular  and  mutually  equilateral,  they  are  congruent. 

130.  The  principal  polygons  are  named  as  follows  : 


No.  OF  Sides 

Name  of  the  Polygon 

No.  OF  Sides 

Name  of  the  Polygon 

3 
4 

6 
6 

Triangle 
Quadrilateral 
Pentagon 
Hexagon 

7 

8 

10 

n 

Heptagon 
Octagon 
Decagon 
w-gon 

QUADRILATERALS  69 

QUADRILATERALS 

131.  A  Parallelogram  (O)  is  a  quadrilateral  whose  opposite 
sides  are  parallel.  

A  pair  of  parallel  sides  are  called  bases;         /  7 

the  perpendicular  distance  between  them  is       / / 

called  the  altitude. 

132.  Cor.     Two  consecutive   angles  of  a  parallelogram  are 
suiyplcmentary. 

For,  in  the  figure  of  §  131,  since  AB  cuts  the  parallels  AD  and  J5(7, 
ZA+ZB=l8\../..     (§103.) 

Ex.  135.     Construct    a    CJ   ABCD,     making  ^  /  g  \ 

AD  =  2  in.,  AB  =  S  in.,  and  B  =  60°.     After  you         /  ~7 

have  constructed  the  figure,  compare  the  opposite      /\\ /3\ 

sides  by  means  of  your  dividers.  e""' 

Proposition  XXIII.     Theorem 

133.  A  diagonal  of  a  parallelogram  divides  it  into 
two  congruent  triangles. 


Hypothesis.     ABCD  is  a  parallelogram.     AC  is  a  diagonal. 

Conclusion.  A  ABC  ^  A  ACD. 

[Proof  to  be  given  by  the  pupil.] 

Suggestions.  — 1.  Since  AD  \\  BC,  compare  Z 1  and  Z  2. 
2.  Compare  Z3  and  Z4.    What  are  the  parallels? 

134.  Cor.  1.     Tlie  opposite  sides  of  a  parallelogram  are  equal. 

135.  Cor.  2.     TJie  opposite   angles  of  a 
parallelogram  are  equal.  ^~7  y     ^ 

136.  Cor.  3.     Segments   of  parallels   in-    Z-^ ^^^ 

eluded  between  parallels  are  equal. 


70  PLANE    GEOMETRY  —  BOOK   I 

Proposition  XXIV.     Theorem 

137.    The  diagonals  of  a  parallelogram  bisect  each 
other. 


Hypothesis.  ABCD  is  a  O. 

Diagonals  AC  and  BD  intersect  at  E. 
Conclusion.  AE  =  EG\  BE  =  ED. 

[Proof  to  be  given  by  the  pupil.] 

Note.  —  The  point  of  intersection  of  the  diagonals  of  a  parallelogram 
is  the  Center  of  the  parallelogram. 

Ex.  136.     If  one  angle  of  a  parallelogram  is  100°,  how  large  is  each  of 
the  other  angles  ? 

Ex.  137.     If  one  angle  of  a  parallelogram  is  a  right  angle,  the  others 
are  also, 

Ex.  138.     If  two  adjacent  sides  of  a  parallelogram  are  equal,  all  its 
sides  are  equal. 

Ex.   139.     Two  parallels  are  everywhere  equidistant. 
Hypothesis.     CD  ||  AB. 

EF  and  GH  are  any  two  Js  to  CD  and  AB. 
Conclusion.     EF  =  GH.  A- 


F  K     " 

Ex.  140.    If  perpendiculars  BE  and  DF  are  drawn 
to  the  diagonal  ^O  of  a  parallelogram  ABCD,  then  BE  =  DF. 
(Construct  the  figure  with  ruler  and  compasses.) 

Ex.  141.  If  a  line  be  drawn  through  the  center  of  a  parallelogram 
and  terminated  by  two  opposite  sides  of  the  parallelogram,  it  is  bisected 
by  the  center. 

Ex.  142.  Construct  the  parallelogram  whose  diagonals  are  2  in.  and 
3  in.  respectively  if  the  included  acute  angle  is  45°.  Measure  the  longer 
and  shorter  sides  of  the  parallelogram. 


QUADRILATERALS  71 

Proposition  XXV.     Theorem 

138.    If  two  sides  of  a  quadrilateral  are  equal  and 
parallel,  the  figure  is  a  parallelogram. 


rmC 


Hypothesis.  AB=CD',  AB  II  CD. 

Conclusion.  ABCD  is  a  parallelogram. 

Plan.     AD  must  be  proved  II  to  BC.    Try  to  prove  Z  1  =  Z  2. 

Proof.     1.  lu  A  ABC  and  A  ACD  : 

AB  =  CD  and  AC  =  AC\  Why  ? 

Z3  =  Z4. 
[Since  lis  ^B  and  CD  are  cut  by  ^C]  Why  ? 

2.  .-.    A  ABC  ^  A  ACD.  Why? 

3.  .-.  Z1  =  Z2.  Why? 

4.  .-.  ^Z>  II  BC  Why  ? 

5.  .-.   ABCD  is  a  parallelogram.  §  131 


B, 


Ex.  143.     The    line    joining    the    mid-points    of 

two  opposite  sides  of  a  parallelogram  is  parallel  to     \^y^ v    /p. 

the  other  two  sides.  /^*^              /^ 

(Prove  AEFD  \s  Hl  O  and  therefore  EF  ||  AD)  ^                       ° 

Ex.  144.     li  ABCD  is  a  parallelogram,  and  E  and  i^are  the  mid- 
points of  AB  and  CD  respectively,  then  AECF  is  also  a  parallelogram. 

Ex.  145.     Prove  that  two  straight  lines  are  parallel  if  any  two  points 
of  one  are  equidistant  from  the  other.     C  Recall  §  84.) 

Ex.  146.     If  the  diagonals  of  a  quadrilateral  bisect  each  other,  the 
figure  is  a  parallelogram, 

Ex.  147.  If  from  any  point  in  the  base  of  an  isosceles 
triangle  parallels  to  the  equal  sides  be  drawn,  the  perimeter 
of  the  parallelogram  formed  is  equal  to  the  sum  of  the 
equal  sides  of  the  triangle.  A^  ^ 

Note.  —  Supplementary  Exercises  50-52,  p.  277,  can  be  studied  now. 


72 


PLANE    GEOMETRY  —  BOOK   I 

Proposition  XXYI.     Theorem 


139.   If  the  opposite   sides  of  a    quadrilateral   are 
equal,  the  figure  is  a  parallelogram. 


2.' 


-^0 


Proof. 
2. 
3. 
4. 

Note.  - 


Hypothesis.  AB=CD;  BC  =  AD, 

Conclusion.  ABCD  is  a  parallelogram. 

Plan.     Try  to  prove  AB  II  CD,  and  AD  II  BC. 

1.       A  ABC  ^  A  ACD.  Give  the  full  proof. 

.-.  Z  1  =  Z  2,  and  hence  BC  II  AD.  Why  ? 

Also  Z  3  =  Z  4,  and  hence  AB  II  CD.  Why  ? 

.-.  .4^  CD  is  aO.  Why? 

Another  proof  may  be  given,  which  is  based  upon  §  138. 

Ex.  148.  If  E,  F,  G,  and  H  are  mid-points  of  sides  AB,  BC,  CD, 
and  AD  respectively  of  parallelogram  ABCD,  then  EFGH  is  a  parallelo- 

^^^"^-  (  Prove  EF  =  HG  and  EH=  FG.) 

Ex.  149.  Construct  a  parallelogram  having  sides  2  in.  and  3  in.  re- 
spectively, and  with  included  angle  of  45°.     Measure  its  longer  diagonal. 

140.    There    is   an    important   applica-       ^. __^^ 

tion  of  parallelograms  in  science.     If  an     ^^/  x^ 

object  is  being  pulled  in  the  direction  AB 

with  a  force  of   50  lb.  and  in  the  direc-    ^  ^^^ 

tion  AC  with  a  force  of  100  lb.,  it  will  actually  move  in  the 

direction  AD  and  as  if  pulled  by  a  force  which  bears  to  100 

lb.  the  same  relation  that  AD  bears  to  AC. 

Thus,  AC=  1"  and  AD  =  11"-,  since  AC  represents  100  lb., 
AD  represents  125  lb. 

Ex.  150.  A  steamer  is  being  propelled  east  at  the  rate  of  15  mi.  an 
hour  ;  the  wind  is  driving  it  north  at  the  rate  of  5  mi.  an  hour.  Deter- 
mine by  a  construction  the  direction  in  which  the  steamer  will  travel  and 
its  rate. 


QUADRILATERALS  73 

SPECIAL  PARALLELOGRAMS 

141.   A  Rectangle  is  a  parallelogram  one  of  whose  angles  is 
a  right  angle.     It  can  be  proved  and  it  is  impor- 
tant to  remember  that  all  the  angles  of  a  rectangle 
are  right  angles.      * 


Note.  —  Since  a  rectangle  is  a  special  parallelogram,  every  theorem 
true  about  parallelograms  is  true  about  rectangles.  Thus,  the  diagonals 
of  a  rectangle  bisect  each  other.  On  the  other  hand,  theorems  true  about 
a  rectangle  are  not  necessarily  true  about  a  parallelogram,  since  a  rectangle 
is  a  special  parallelogram. 

Ex.  151.  State  some  other  properties  of  a  rectangle  which  follow  at 
once  from  properties  of  a  parallelogram.     (See  §§  133-137.) 

Ex.  152.  Construct  a  rectangle  whose  sides  are  1.5  in.  and  2  in. 
respectively.     Draw  and  measure  its  diagonals. 

Ex.  153.     Prove  that  the  diagonals  of  a  rectangle  are  equal. 

Ex.  154.  Prove  that  a  quadrilateral  whose  angles  are  all  right  angles 
is  a  rectangle. 

Ex.  155.  Prove  that  a  parallelogram  whose  diagonals  are  equal  is  a 
rectangle.  B 

Plan.    Try  to  prove  one  of  its  ^i  is  a  right  angle. 

(Recall  §  116,  G  1,  and  also  §  103.) 


Ex.  156.  When  laying  out  the  lines  for  the  foundation  of  a  rec- 
tangular building,  as  ABCD,  contractors  often  measure  off  AD  and  DC 
at  riglit  angles  and  of  the  required  lengths.  Then  AB  is  measured  off 
equal  to  CD  and  at  right  angles  to  AD.     (See  figure  for  Ex.  155.) 

(a)  Why  should  ABCD  then  be  a  rectangle  ? 

(6)  To  test  whether  ABCD  is  a  true  rectangle,  AG  and  BD  are 
measured.  If  they  prove  to  be  equal,  it  is  concluded  that  the  figure  is  a 
rectangle.     Is  this  a  safe  test  ?     Why  ? 

Note.  —  Supplementary  Exercises  53-64,  p.  278,  can  be  studied  now. 

142.   A  Rhombus  is  a   parallelogram   having  two  adjacent 

sides  equal.     It  can  he  proved  and  it  is  important         / 7 

to  remember  that  all  the  sides  of  a  rhombus  are       /  / 

equal ;  also  it  is  usually  implied  that  the  angles    Z / 

are  not  right  angles.     (See  §  143.) 

Ex.  157.  State  properties  of  a  rhombus  which  are  evident  at  once 
because  the  rhombus  is  a  special  parallelogram.     (See  Note,  §  141.) 


74 


PLANE    GEOMETRY  —  BOOK   I 


Ex.  158.  Prove  that  the  diagonals  of  a  rhombus  are  perpendicular 
to  each  other. 

Ex.  159.     Prove  that  the  diagonals  of  a  rhombus  bisect  the  angles. 

Note.  —  Supplementary  Exercises  55-57,  p.  278,  can  be  studied  now. 

143.  A  Square  is  a  parallelogram  having  two  adjacent  sides 
equal  and  one  angle  a  right  angle.  It  can  he  proved  and  it  is 
important  to  remember  that 

All  the  angles  of  a  square  are  right  angles  and  all  the  sides  are 
equal. 

Note.  —  The  square  is  a  special  rectangle  and  also  a  special  rhombus. 
Hence  every  theorem  true  about  a  rectangle  or  a  rhombus  is  true  about  a 
square.     (See  Note,  §  141.) 

144.  Many  artistic  designs  are  made  on  a  network  of  squares 
as  illustrated  below. 


!"S 

SI 

Iri: 

"SP 

H"ss:s 

H«.. 

.HHH. 

ir-!:- 

in. 

i-f:.! 

l:i 

is!:! 

:g; 

Bs'nil 

t^^• 

m\n 

^■■■■■■■■■■niSaai 

id>« 

■   ■ 
SS  ■ 

1 

1 

1 

«,7,t 

[hhh. 

jgS  hi  S.!  ij  !i  li 

loo 

M 

Corner  and  Border 


Four  Border  Designs 


Rug  Designs 


QUADRILATERALS  75 

Ex.  160.  Make  a  list  of  facts  about  the  square  which  may  be  inferred 
from  known  facts  about  the  parallelogram,  the  rectangle,  and  the 
rhombus. 

Ex.  161.  How  large  are  the  angles  into  which  a  diagonal  of  a  square 
divides  its  angles  ? 

Ex.  162.     Construct  a  square  whose  diagonals  shall  be  2  in.  in  length. 

Ex.  163.  Prove  that  the  lines  drawn  from  the  ends  of  one  side  of  a 
square  to  the  mid-points  of  the  two  adjacent  sides  are  equal. 

Ex.  164.  Prove  that  if  the  diagonals  of  a  quadrilateral  ate  perpen- 
dicular to  and  bisect  each  other,  the  figure  is  a  rhombus. 

Ex.  165.  If  E,  F,  G,  and  H  are  points  on  the  sides,  AB,  BC,  CD, 
and  AD  respectively  of  square  ABCD,  such  that  AE  =  BF  =  CG  =  DH, 
prove  that  EFGH  is  a  square. 

Suggestions.  — 1.  Try  to  prove  EFGH  i&  a  O,  having 
two  adj.  sides  equal,  and  having  one  Z{Z^)  a  right  angle. 
(§  143.) 

2.   To  prove  Z4  a  right  angle: 

(a)Zl+Z2=?  (6)  DoesZ3  =  Z2? 

(c)  2:i  +  Z3  +  Z4=  ?  (d)  .■.Zi=  ? 

Note.  —  Supplementary  Exercises  58-60,  p.  278,  can  be  studied  now. 

TRAPEZOIDS 

145.  A  Trapezoid  is  a  quadrilateral  which  has  one  and  only 
one  pair  of  parallel  sides  ;  AB  and  CD  are  ^ 

called  the  non-parallel  sides.  /  \ 

The  parallel  sides  of  a  trapezoid  are  called  the       /                       \ 
Bases.  ^^ ^^ 

The  perpendicular  distance  between  the  bases  is  called  the  Altitude. 

The  line  joining  the  mid-points  of  the  non-parallel  sides  is  called  the 
Median  of  the  trapezoid. 

146.  An  Isosceles  Trapezoid  is  a  trapezoid  the  non-parallel 
sides  of  which  are  equal. 

Ex.  166.  Construct  the  trapezoid  having  lower  base  of  4  in.,  one  of 
its  non-parallel  sides  2  in.,  the  angle  between  these  two  sides  being  60°, 
and  the  upper  base  being  1.6  in. 

Ex.  167.  If  the  angles  at  the  ends  of  one  base  of  a  trapezoid  are 
equal,  the  angles  at  the  ends  of  the  other  base  are  also  equal. 


76  PLANE    GEOMETRY  —  BOOK   I 

Ex.  168.     If  a"  trapezoid  is  isosceles,  the  lower  base  angles  are  equal. 
(If  AB=  CD,  prove  ZA  =  ZD.     Draw  BE  ||  CD.  g 

Compare  Z  AEB  with  Z  D  and  Z  A.)  A  \ 

Ex.  169.     If  one  pair  of  base  angles  of  a  trape-        /      \  \ 

zoid  are  equal,  the  trapezoid  is  isosceles.  ^  E  ^ 

Ex.  170.     Prove  that  the  diagonals  of  an    isosceles    trapezoid   are 
equal. 

Ex.  171.   Prove  that  the  opposite  angles  of  an  isosceles  trapezoid  are 

supplementary. 

Note.  —  Supplementary  Exercises  61-63,  p.  278,  can  be  studied  now. 


Proposition  XXYII.     Theorem 
147.  If  three  or  r)%ore  parallels  intercept  equal  lengths 
on  one  transversal,  they  intercept  equal  lengths  on  all 
transversals. 


Al 

\B 

\L 

)ih\D 

M 

r^l^F 

Gl 

kf\H 

Hypothesis.  AB  !1  CD  ||  EF  II  GH. 

AG  cuts  the  lis  at  A,  C,  E,  and  G. 
BH  cnts  the  lis  at' 5,  D,  F,  and  H. 
AC=CE  =  EG. 
Conclusion.  BD  =  DF=  FH. 

Plan.     Try  to  prove  BD,  DF,  and  FH  homologous  sides  of 
cong.  A. 

Proof.    1.       Draw  BI,  DJ,  F/f  parallel  to  AG. 

§91 

EG.  Why  ? 

Hyp. 

Ax.  1,  §  51 

[Complete  the  proof  by  proving  ^  jBZ)/,  DJF,  and  FHK 
are  congruent,  and  then  proving  that  BD  =.  DF  -—  FH.'] 


2 

.'.BIWDJWFK 

3. 

Also 

BI=AC,  DJ=  CE,  and  FK 

4. 

But 

AC=CE=EG. 

5. 

.-.  BI=DJ=FK. 

QUADRILATERALS 


77 


148.   Cor.  1.     If  a  line  bisects  one  side  of  a  triangle,  and  is 
parallel  to  a  second  side,  it  bisects  the  third  side  also. 

Hyp.  I)  is  on  AB  of  A  ABC  ; 

AD  =  DB;  DEWBG. 
Con.  AE  =  EC. 

Proof.   1.  Assume  X4r  II  ^C 


'K 

c 

f::A 

\ 

149.  Cor.  2.     If  a  line  is  parallel  to  the 

bases  of  a  trapezoid  and  bisects  one  of  the 

non-parallel  sides,  it  bisects  the  other  also. 

B    '  O 

Note.  —  Supplementary  Exercises  64-66,  p.  278,  can  be  studied  now. 

Proposition   XXVIII.     Problem 

150.  Divide   a  given  segment  into  any  number  of 
equal  parts,  ^ 


Ct)(< 


^V< 


I     '1 


P\' 


I 
I 

I 


"--^.v 


IF 


Given  segment  AB. 

Required  to  divide  AB  into  five  equal  parts. 
Construction.    1.    Draw   line   AC,    making   a   convenient  Z 
with  AB 

2.  Upon  AC,  lay  off  AD=DE  =  EF=  FG  =  GIL 

3.  Draw  IIB. 

4.  Through  D,  E,  F,  G,  and  H,  draw  lines  parallel  to  IIB, 
meeting?  .17?  at  X,  Y,  Z,  and  W. 

Statement.       AX=XY=^YZ=Z  W=  WB. 
Proof.    1.    Assume  RS  through  A  parallel  to  HB. 

2.  .'.  RSW  DXWEYWFZWGWWHB.  Why? 

3.  .-.  AX=XY=  YZ=  ZW=  WB.  Why? 

Note.  —  Supplementary  Exercises  66-67,  p.  279,  can  be  studied  now. 


78  PLANE    GEOMETRY  —  BOOK   I 

Proposition  XXIX.     Theorem 

151.  If  a  line  joins  the  midpoints  of  two  sides  of  a 
triangle,  it  is  parallel  to  the  third  side  and  equal  to 
one  half  of  it. 


Hypothesis.  D  is  the  mid-point  of  AB,  and  E  is  the  mid- 
point of  AC  in  A  ABC. 

Conclusion.  DE  W  BG;  DE  =  \  BC 

Plan.  Extend  DE  its  own  length  to  F.  Try  to  prove 
FE  =  BC,  and  FE  II  BG.  To  do  this,  try  to  prove  FECB 
is  a  lJ. 

Proof.     1.    Extend  DE  to  F,  making  DF=  DE.     Draw  BF. 

2.  .-.A  FBD  ^  A  DAE.  Give  the  proof. 

3.  .-.  Z  1  =  Z  2  ;  and  also  BF  =AE.  Why  ? 

4.  .-.  BF  II  AC,  and  .-.  BF  II  EC.  Why  ? 

5.  AhoBF=Ea  Why? 

6.  .-.  ^2^£;(7  is  a  parallelogram.         -  Why  ? 

7.  .-.  i^^  or  DE  is  parallel  to  ^C.  Why  ? 

8.  Aho  FE  =  BC,  3ind  .:  DE  =  i  BC.  Why? 
Note.  — This  theorem  is  very  important. 

152.  The  proof  of  Proposition  XXIX  illustrates  another 
valuable  device  for  proving  theorems. 

Principle  III.  To  prove  that  one  segment  is  double  another, 
either  double  the  shorter  and  prove  the  result  equal  to  the 
longer,  or  halve  the  longer  and  prove  the  result  equal  to  the 
shorter.  The  first  of  these  plans  is  followed  in  the  proof  of 
Proposition  XXIX;  the  second  plan  will  be  used  in  Propo- 
sition XL. 


QUADRILATERALS  79 

Ex.  172.  The  lines  joining  the  mid-points  of  the  sides  of  a  triangle 
divide  it  into  four  congruent  triangles. 

Ex.  173.  If  J57,  F,  G,  and  ^are  the  mid-points  of  the  sides  AB,  BC, 
CD,  and  AD  respectively  of  a  quadrilateral  ABCD,  then  EFGH  is  a 
parallelogram.  (Draw  AC  and  use  Proposition  XXIX.)  This  theorem 
appeared  in  a  book  on  geometry  by  Th.  Simpson  in  1760.         a 

Ex.  174.  The  lines  joining  the  mid-points  of  the  oppo- 
site sides  of  a  quadrilateral  bisect  each  other. 

Ex.  175.  The  mid-point  of  the  hypotenuse  of  a  right 
triangle   is   equidistant  from  the  vertices  of  the   triangle. 

(Let  AE  =  EB.  Vroxe  ED  ±AB.  Then  complete 
the  proof. )  B 

Proposition  XXX.     Theorem 

153.  The  median  of  a  trapezoid  is  parallel  to  the 
bases  and  equal  to  one  half  their  sum, 

A  B 


52^_^::_^:^., 


Hypothesis.  ABCD  is  a  trapezoid. 

E  is  the  mid-point  of  AD  and  F  of  BC. 
Conclusion.  .    EF  II  AB  and  DC. 

EF=i(AB-\-DC). 
Proof.    1.   Extend  DC  to  G,  making  CO  =  AB.    Draw  AC, 
BG,  and  AG. 

2.  .'.ABGCissLCJ.  Why? 
[Since  CG  =  AB,  and  CG  II  AB.-\ 

3.  .-.  AG  passes  through  i^and  is  bisected  by  it.       §  137 

4.  .-.  in  A  ADG,  AE  =  ED  and  AF  =  FG. 

5.  .-.  EF  II  DG ;  and  EF  =  i;  DG.  Why  ? 

6.  .'.  EF  W  DC  B,nd  AB. 

7.  Also  EF  =  i  (DC  4-  AB),  since  DG=  DC-\-  AB. 

Note.  —  Supplementary  Exercises  68-74,  p.  279,  can  be  studied  now. 


80 


PLANE    GEOMETRY  —  BOOK   I 


Proposition  XXXI.     Theorem 
154.    The  sum  of  the  interior  angles  of  a  i^olygon 


having  n  sides  is  (n  —  2)  straight  angles. 


Hypothesis.     Assume  a  polygon  of  n  sides,  like  ABCD  ••• . 
Conclusion.     The  sum  of  its  int.  A={n  —  2)  st.  A. 
Proof.     1.    Draw   diagonals   from  B  to  each  of   the   other 
vertices. 

2.  Each  side  of  the  polygon,  excepting  AB  and  BC,  becomes 
the  base  of  a  triangle  whose  vertex  is  at  B.  Hence  there  are 
(n  —  2)  A  formed. 

That  is,  when  n  is  4,  there  are  2  A  ; 
when  w  is  5,  there  are  3  A  ; 
when  n  is  6,  there  are  4  A ;  etc. 

3.  The  sum  of  the  int.  A  of  each  A  is  1  st.  Z.       Why  ? 

4.  .  • .  the  sum  of  the  int.  A  of  the  (n  —  2)  A  is  {n  —  2)  st.  A. 

5.  But  the  sum  of  the  int.  A  of  the  A  =  the  sum  of  the 
int.  A  of  the  polygon. 

6.  .*.  the  sum  of  the  int.  A  of  the  polygon  is  (n  —  2)  st.  A. 

Ex.  176.  Express  in  straight  angles,  in  right  angles,  and  in  degrees 
the  sum  of  the  angles  of  a  polygon  having : 

{a)  four  sides  ;  (6)  five  sides  ;  (c)  six  sides ;  (d)  eight  sides. 

Ex.  177.  How  many  degrees  are  there  in  each  angle  of  an  equi- 
angular : 

{a)  quadrilateral?     (6)  pentagon?     (c)  hexagon?     {d)  octagon? 

Ex.  178.  If  two  angles  of  a  quadrilateral  are  supplementary,  then 
the  other  two  are  also. 

Ex.  179.  How  many  sides  has  a  polygon  the  sum  of  whose  angles  is 
16  right  angles  ?    7  straiglit  angles  ?     1620  degrees  ? 


POLYGONS  81 

Proposition  XXXII.     Theorem 

155.  If  the  sides  of  any  polygon  he  extended  in  order 
to  form  an  exterior  angle  at  each  vertex,  the  sum  of 
these  exterior  angles  is  two  straight  angles. 


Hypothesis.     Assume  a  polygon  of  n  sides. 

Extend  the  sides  as  in  the  figure. 
Conclusion.     The  sum  of  ext.  A  like  Z  1,  Z  2,  Z  3,  etc.  = 

2  St.  A. 
Proof.     1.   The  sum  of  the  int.  Z  and  the  ext.  Z  at  each 
vertex  =  1  st.  Z.  §  39 

2.  .'.  the  sum  of  all  the  int.  and  ext.  A  =  n  st.  A.      Why? 

3.  But  the  sum  of  all  the  int.  Z  =(?i  -  2)  st.  A.        §  154 

4.  .*.  the  sum  of  all  the  ext.  Z  =  2  st.  A. 

Note.  —  Propositions  XXXI  and  XXXII  were  proved  in  their  general 
form  by  Regiomontanus  (1436-1476),  although  the  theorems  were  known 
to  earlier  mathematicians  and  were  proved  by  them  for  special  cases. 

Ex.  180,     Prove  the  theorem  of  §  154  by  drawing 
lines  from  any  point  within  the  polygon  to  the  vertices.       ^,^^'''T^"""~~--^ 
(Recall  §  .35.)  ^--  i--'^ 

Ex.  181.     State  and  prove  the  converse  of  §  135.  \      /  '  n      / 

Suf/f/estion.  —  Apply  §  154  and  §  98.  \/''  -J 

Ex.  182.     How  many  sides  are  there  in  the  polygon 
the  sum  of  whose  interior  angles  exceeds  the  sum  of  its  exterior  angles 
by  540^^  ? 

Ex.  183.  How  many  sides  has  a  polygon  the  sum  of  whose  interior 
angles  equals  four  times  the  sum  of  its  exterior  angles  ? 


82  PLANE    GEOMETRY  —  BOOK   I 

INEQUALITIES 

156.  The  symbol  for  "  less  than  "  is  <  ;  for  ^^  greater  than  " 
is  >. 

157.  Order  of  Inequalities,  a  <h  and  c  <  d  are  two  in- 
equalities of  the  same  order,  m  <  w  and  x  ^  y  are  two  in- 
equalities of  opposite  orders. 

158.  Axioms  for  combining  Inequalities. 

Ax.  17.  If  equals  be  added  to  unequals,  the  sums  are  unequal 
in  the  same  order. 

Thus,  if  a  <  &,  then  a  +  c  <Cb  ■{■  c. 

Ax.  18.  If  equals  he  subtracted  from  unequals,  the  differences 
are  unequal  in  the  same  order. 

Thus,  if  a  <  6,  then  a  —  c<ih  —  c. 

Ax.  19.  If  unequals  be  added  to  unequals  in  the  same  order, 
the  sums  are  unequal  in  the  same  order. 

Thus,  if  a  <  6,  and  c  <  (?,  then  a  +  c  <  6  -1-  d. 

Ax.  20.  If  unequals  be  subtracted  from  equals  or  from  un- 
equals of  opposite  order,  the  differences  are  unequal  and  of  order 
opposite  to  that  of  the  subtrahend. 

Thus,  if  a  >  6,  and  c  <  <?,  then  a  —  c  >  6  —  d. 
Arithmetical  Example.  —  Since  12  >  7  and  3  <  5,  then  12  —  3  should 
be  greater  than  7  -  5.     Is  it  ? 

Ax.  21.     Ifa>b  and  b  >  c,  then  a  >  c. 

Ex.  184.     Given  an  arithmetical  example  for  each  of  the  axioms. 

159.  Fundamental  Inequalities  for  Segments. 

(a)  Any  side  of  a  triangle  is  less  thari  the  sum 
of  the  other  two  sides. 

This  follows  from  Ax.  11,  §51. 
Thus  BC<AB-\-AC. 


INEQUALITIES  83 

(b)  Any  side  of  a  triangle  is  greater  than  the 
difference  between  the  other  two  sides. 

Thus,  BOAC-AB. 

For,  from  (a)  BC  +  AB  >  AC.     Subtracting  AB 
from  both  members  of  the  inequality,  BO  AC—  g 
AB,  by  Ax.  18,  §  158. 

Note. — Ex.  188,  p.  86,  and  Supplementary  Exercises  75-84,  p.  280, 
can  be  studied  now. 

160.  Fundamental  Inequality  for  Angles. 

An  exterior  angle  of  a  triangle  is  greater  than  either  remote 
interior  angle  of  the  triangle.     (§  87.) 

Proposition  XXXIII.     Theorem 

161.  If  hvo  sides  of  a  triangle  are  unequal,  the 
angles  opposite  are  unequal,  the  angle  oppjosite  the 
greater  side  being  the  greater. 


Hypothesis.  In  A  ABC,  AC  >  AB. 

Conclusion.  ZB  >  Z.C. 

Proof.  1.    Since  AC  >  AB,  take  AD  =  AB.     Draw  BD. 

§  13 

2.  .-.  Z1=Z2.  Why? 

3.  Z  2  is  an  exterior  angle  of  A  BDC.  Def. 

4.  .-.  Z2>  Z  C.  Why? 

5.  .-.  Z1>ZC.  Why? 

6.  But  ZABC>Z1.  Ax.  8,  §51 

7.  .-.  Z  ABC  >  ZC  Ax.  21,  §  158 

Ex.  185.    If  a  triangle  is  scalene,  all  its  angles  are  unequal. 


84 


PLAJNTE    GEOMETRY  —  BOOK   I 


Proposition  XXXIV.     Theorem 

162,  If  tivo  angles  of  a  triangle  are  uJiequaL  the 
sides  opposite  are  uneqiml,  the  side  opposite  the  greater 
angle  being  the  greater. 


Hypothesis.  Jn  A  ABC,  Z.C  <  Z  B. 

Conclusion.  AB  <  AC. 

Proof.  1.    Since  Z  B  >  Z  C,  construct  BD,  making  Z1=ZC. 

2.  .'.BD  =  DC.  §123 

3.  In  A  ABD,  AB  <  AD -{-  BD.  §  159,  a 

4.  .-.  AB<AD  +  DC,  or  AB  <  AC. 
[Substitute  DC  for  its  equal,  DB.] 

163.  Cor.  1.  The  hypotenuse  of  a  right  tri- 
angle is  greater  than  either  leg  of  the  triangle. 

164.  Cor.  2.  The  perpendicular  from  a  point 
to  a  line  is  the  shortest  segment  from  the  point  to 
the  line. 

165.  Cor.  3.  If  two  oblique  .segments, 
drawn  from  a  point  in  a  perpendicular 
to  a  line,  cut  off  unequal  distances  from 
the  foot  of  the  perpendicular,  the  more 
remote  is  the  greater. 


Hyp.     CD±AB;  ED>  BF.        Con.     CE  >  CF. 

Suggestions.  — 1.  Take  DH=  BF,  and  draw  CH. 

2.  Prove  CH=  CF. 

3.  Prove  Z  2  >  Z 1,  by  comparing  eacb  with  Z  3. 

4.  Then  complete  the  proof. 


% 


\ 
^    ^ 


i 


INEQUALITIES 


85 


Ex.  186.  If  0  is  any  point  in  the  base  BC  oi  isosceles  triangle  ABC^ 
then  AO  is  less  than  AC.     (Prove  Z^OOZ^CO.) 

Ex.  187.  Prove  that  the  median  to  any  side  of  a  triangle  is  greater 
than  the  altitude  to  that  side  unless  the  side  is  the  base  of  an  isosceles 
triangle. 

Proposition  XXXV.     Theorem 

166.  If  two  triangles  have  tioo  sides  of  one  equal 
respectively  to  two  sides  of  the  other,  hut  the  included 
angle  of  the  first  greater  than  the  included  angle  of  the 
second,  then  the  third  side  of  the  first  is  greater  than 
the  third  side  of  the  second. 

D 


Hypothesis.  In  A  ABC,  and  A  DEF: 

AB=DE',  AC=DF',  /.BAOZ.D. 
Conclusion.  BC  >  EF. 

Proof.     1.   Place  A  DEF  in    the   position  ABG,  side  DE 
coinciding  with  its  equal  AB. 

2.    DF  falls  within  Z  BAC,  taking  the  position  AG.  Why  ? 
3*.   Construct  AH  bisecting  Z  GAC,  and  meeting  BC  at  H. 
Draw  GH. 

A  GAH  ^  A  ACIL        Give  the  full  proof. 


.-.  GH=  CH. 

In  A  BUG,  BH+  GH>BG. 

BH+CH>  BG,  or  BC  >  BG. 


Note.  —  If  (r  falls  on  BC,  then  EF  is  at  once  less  than  BC. 
within  A  ABC,  the  proof  is  similar  to  that  given  in  the  text. 


Why? 

Why? 

Wliy  ? 

If  G  falls 


86 


PLANE    GEOMETRY  —  BOOK   I 


Proposition  XXXYI.     Theorem 

167.  If  two  triangles  have  two  sides  of  one  equal 
respectively  to  two  sides  of  the  other,  hut  the  third  side 
of  the  first  greater  than  the  third  side  of  the  second, 
then  the  angle  opposite  the  third  side  of  the  first  is 
greater  than  the  angle  opposite  the  third  side  of  the 
second. 


B 
Hypothesis.  In  A  ABC  and  A  I)EF: 

AB  =  DE,  AC  =  BF',  BOEF. 
Conclusion.  ^A>Z  B. 

Proof.     1.    Suppose  that  Z  Ais  not  greater  than  Z  D  ;  that 
is,  that  Z  A  either  equals  Z  D  or  is  less  than  Z  D. 


2. 


li  ZA 


Why? 

Hyp. 

§94 

§166 

Hyp. 


Z  B,  then  A  ABC  ^  A  BEF. 
[Give  the  full  proof.] 

ThenBC=EF. 
But  BC>  EF. 
.-.  Z  A  cannot  be  equal  to  Z  B. 
liZA<ZB,  then  BC  <  EF. 
But  BC  >  EF. 
.-.  Z  A  cannot  be  less  than  Z  B. 
Since  Z  A  cannot  be  equal  to  Z  B  or  less  than  Z  B,  then 
Z  A  must  be  greater  than  Z  B. 

Ex.  188.     If  0  is  any  point  within  A  ^J5C,  then  ^0+0(7  <^^+^ (7. 
Suggestions.  —  1.   Extend   CO  until  it  intersects 
AB  at  R.    2.   Compare  BO  with  BR  +  RO. 

3.  Add  OC  to  both  members  of  the  inequality. 

4.  Compare  RO+OC  with  RA-jrAC,  and  com-        ^ — a^^ 

plete  the  proof. 


SUPPLEMENTARY  THEOREMS  87 

SUPPLEMENTARY  THEOREMS 

168.  Three  or  more  lines  ordinarily  do  not  pass  through  a 
common  point.  Three  or  more  lines  which  do  pass  through  a 
common  point  are  called  Concurrent  Lines. 

Proposition  XXXVII.     Theorem 

169.  The  bisectors  of  the  interior  angles  of  a  tri- 
angle meet  at  a  point  ivhich  is  equidistant  from  the 
sides  of  the  triangle. 


Hypothesis.  In  A  ABC : 

AD  bisects  Z^  A,  BE  bisects  Z  B]  Ci^ bisects  Z  C. 
Conclusion.     AD,  BE,   and    CF  meet  at  a  point  which   is 
equidistant  from  the  sides  of  A  ABC. 

Proof.     1.   Let  AD  and  BE  meet  at  point  P.  Note  1 

2.  Since  P  is  in  AD,  it  is  equidistant  from  AC  and  AB. 

§  120,  I 

3.  Since  P  is  in  BE,  it  is  equidistant  from  AB  and  BG. 

4.  .-.  P  is  equidistant  from  AC  and  BC.     Ax.  1,  §  51 

5.  .-.  P  lies  in  CF,  the  bisector  of  Z  C.  §  120,  II 

6.  Hence  AD,  BE,  and  CF  meet  at  P,  which  is  equidistant 
from  AB,  AC,  and  BC. 

Note  1. — This  fact  may  be  assumed  as  evident  from  the  figure,  or 
may  be  proved  as  follows  : 
-    1.   If  AD  does  not  intersect  BE,  then  AD  II  BE. 

2.  Then  Z  DAB  +  Z  EBA  =  1  St.  Z.     (§  103.)      ' 

3.  But  this  is  impossible,  since  Z  DAB  +  Z  EBA  <  1  st.  Z.        Why  ? 

Note  2.  —  The  point  of  intersection  of  the  bisectors  of   the  interior 
angles  of  a  triangle  is  called  the  In-center  of  the  triangle.     (See  §  226.) 


88  PLANE    GEOMETRY  —  BOOK   I 

Proposition  XXXVIII.     Theorem 

170.  The  perpendicular-bisectors  of  the  sides  of  a  tri- 
angle r)ieet  at  a  point  which  is  equidistant  from  the 
vertices  of  the  triangle. 


Hypothesis.  In  A  ABC,  FK,  DO,  and  EH  are  the  perpen- 
dicular-bisectors of  AB,  BC,  and  AC,  respectively. 

Conclusion.  FK,  DO,  and  EH  meet  at  a  point  which  is 
equidistant  from  A,  B,  and  C. 

Proof.     1.    Let  FK  and  DO  meet  at  point  0.  Note  1 

2.  Since  0  is  in  FK,  O  is  equidistant  from  A  and  B.       Why  ? 

3.  Since  0  is  in  DO,  0  is  equidistant  from  B  and  C      Why  ? 

4.  .•.  0  is  equidistant  from  A  and  C.  Ax.  1,  §  51 

5.  .-.0  lies  in  EH,  or  ^^  passes  through  0.  §  118 

6.  Hence  the  perpendicular-bisectors  are  concurrent  at  a 
point  which  is  equidistant  from  A,  B,  and  C 

Note  1.  — This  fact  may  be  assumed  or  be  proved  as  follows  : 

1.  If  i^^does  not  intersect  GD,  then  FKW  GD. 

2.  .-.  AB,  which  is  ±  to  FK,  is  also  ±  to  GD. 

3.  But  BD  ±  GD. 

4.  .-.  either  AB  11  jBZ>,  or  AB  coincides  with  BD. 

5.  But  this  is  impossible,  since  AB  and  BD  intersect. 

Note  2.  —  The  point  of  intersection  of  the  perpendicular-bisectors  of  the 
sides  of  a  triangle  is  called  the  Circum-center  of  the  triangle,  for  a  circLe 
can  be  drawn  with  il  as  center  which  will  pass  through  the  vertices  of  the 
triangle. 

Ex.  189.  Construct  a  circle  which  will  pass  through  the  vertices  of 
the  triangle  the  sides  of  which  are  3  in.,  3  in.,  and  4  in.,  respectively. 


SUPPLEMENTARY  THEOREMS 


89 


Proposition  XXXIX.     Theorem 
171.    The  altitudes  of  a  triangle  meet  at  a  point. 


Hypothesis.  In  A  ABC^  AD,  BE,  and  CF  are  the  altitudes 
from  A,  B,  and  C  respectively. 

Conclusion.     AD,  BE,  and  CF  meet  at  a  point. 

Proof.  1.  Drawjy/rthrough^llto^C;  7t 6^  through  5  II  to 
AC;  and  GH  through  C II  AB.     These  parallels  form  A  HKG. 

2.  Since  AD  1  BC,  then  AD  ±  HK.  Why  ? 

3.  KACB  and  ABCII  are  [U.  Why  ? 

4.  KA  =  ^(7,  and  ^£^  =  -BO.  Why  ? 

5.  .'.  /f^  =  AH,  and  ^Z)  is  the  perpendicular-bisector  of  KH. 

6.  Similarly  BE  and  02^  can  be  proved  to  be  the  perpendicu- 
lar-bisectors of  KG  and  GH  respectively. 

7.  .-.  AD,  BE,  and  CF  in  A  HKG  meet  at  a  point.        §  170 

Note. — The  point  of  intersection  of  the  altitudes  of  a  triangle  is 
called  the  Ortho-center  of  the  triangle. 

Ex.  190.  If  /is  the  ortho-center  (Note,  §  171)  and 
J  is  the  circum-center  (Note,  §  170)  of  triangle  ABC^ 
then  BI=2JK  and  AI=2  JL. 

Suggestions.  —1.  Recall  §  152. 

2.  Prove  3/iV^II  KL,  BIW  JK,  and  AIW  JL. 

3.  Recall  §105. 

Ex.  191.  Does  the  ortho-center  of  a  triangle  necessarily  fall  inside 
the  triangle  ? 

Note.  —  Supplementary  Exercise  86,  p.  281,  can  be  studied  now. 


90  PLANE    GEOMETRY  —  BOOK   I 

Proposition  XL.     Theorem 

172.  The  medians  of  a  triangle  meet  at  a  point 
which  lies  tivo  thirds  the  distance  from  each  vertex  to 
the  mid-point  of  the  opposite  side, 

C 


Hypothesis.     AD,  BE,  and  CF  are  the  medians  of  A  ABC. 

Conclusion.  AD,  BE,  and  CF  meet  at  a  point  which  lies 
two  thirds  the  distance  from  each  vertex  to  the  mid-point  of 
the  opposite  side. 

Proof.     1.    Let  AD  and  BE  meet  at  point  0.     Note  1,  §  169. 

2.  Let  Gf  and  H  be  the  raid-points  of  AG  and  BO  respec- 
tively.    Draw  ED,  GH,  EG,  and  DH. 

3.  Then,  in  A  AOB,  GH^^AB  and  GH  II  AB.        Why  ? 

4.  Similarly,  ED  =  \AB  and  ED  II  AB, 

5.  .'.EDHG  is  2i  a.  Why? 

6.  .-.  GD  and  EH  bisect  each  other.  Why  ? 

7.  .-.  OD=OG  =  AG,  and  EO=OH=  HB. 

8.  HcQce  AD  and  BE  meet  at  a  point  which  lies  two  thirds 
the  distance  from  ^  to  D  and  from  B  to  E. 

9.  In  like  manner,  AD  and  CF  meet  at  a  point  which  lies 
two  thirds  the  distance  from  A  to  D  and  from  C  to  F.  On 
AD,  this  is  point  0. 

10.  Hence  the  three  medians  meet  at  point  0,  which  is  two 
thirds  the  distance  from  each  vertex  to  the  mid-point  of  the 
opposite  side. 

Note.  —  The  point  of  intersection  of  the  medians  of  a  triangle  is  called 
the  Center  of  Gravity  of  the  triangle. 

This  theorem  was  known  to  Archimedes. 


MISCELLANEOUS  EXERCISES  91 

Exercises  Solved  by  Indirect  Proofs 
Ex.  192.     If  two  straight  lines  are  cut  by  a  transversal,  and  a  pair  of 
alternate  interior  angles  are  unequal,  the  lines  are  not  parallel. 
Suggestion.  —  Review  §  94. 

Ex.  193.     If  two  lines  are  cut  by  a  transversal  and  the  sum  of  the 

interior  angles  on  the  same  side  of  the  transversal  is  not  equal  to  two 
right  angles,  the  lines  are  not  parallel. 

Ex.  194.  If  a  point  is  unequally  distant  from  the  ends  of  a  segment, 
it  is  not  in  the  perpendicular-bisector  of  the  segment. 

Ex.  195.  If  a  point  is  not  equidistant  from  the  sides  of  an  angle,  it  is 
not  in  the  bisector  of  the  angle. 

Ex.  196.  Prove  that  the  two  altitudes  of  a  parallelogram  which  has 
two  unequal  sides  are  unequal. 

Ex.  197.  If  two  unequal  oblique  segments  be  drawn  from  a  point  to  a 
straight  line,  the  greater  cuts  off  the  greater  distance  from  the  foot  of  the 
perpendicular  from  the  point  to  the  line. 

Suggestion. — Recall  §  165. 

Miscellaneous  Exercises 

Ex.  198.  If  D,  E,  and  JP'are  points  on  the  sides  AB, 
BC\  and  AC  respectively  of  equilateral  triangle  ABC^ 
such  that  AD  =  BE  =  CF,  then  A  DEF  is  also  equilat- 
eral. '"'"T 

Ex.  199.  Prove  that  the  bisectors  of  a  pair  of  vertical  angles  form  a 
straight  line. 

Ex.  200.  If  two  lines  are  cut  by  a  transversal  so  that  a  pair  of  exterior 
angles  on  the  same  side  of  the  transversal  are  supplementary,  the  lines  are 
parallel. 

Ex.  201.  If  perpendiculars  be  drawn  to  the  sides  of  an  acute  angle 
from  a  point  outside  of  the  angle,  they  form  an  angle  equal  to  the  given 
angle. 

Ex.  202.  If  through  any  point  D  in  one  of  the  equal 
sides  AB  of  isosceles  A  ^-BC,  DFhe  drawn  perpendicular 
to  base  BC,  meeting  CA  extended  at  E,  then  A  ADE  is 
isosceles. 

SuqgeHtion.  —  Compare  LE  with  Z  C,  and  LBDF  vj'Mh 
LB.'  B 

Ex.  203.  Prove  that  the  altitudes  drawn  to  homologous  sides  of  con- 
gruent triangles  are  equal. 


92 


PLANE    GEOMETRY  —  BOOK   I 


Ex.  204.  If  D  is  mid-point  of  side  BG  of  A  ABC, 
and  BE  and  CF  are  perpendiculars  from  B  and  C  to 
AD,  extended  if  necessary,  prove  BE  =  CF. 

Ex.  205.     If  a  line  be  drawn  through  the  vertex  of 

an  isosceles  triangle  parallel  to  the  base,  it  bisects  the 
exterior  angle  at  the  vertex. 

Ex.  206.  Prove  that  the  segments  bisecting  the  base  angles  of  an 
isosceles  triangle  and  terminating  in  the  opposite  sides  are  equal . 

Ex.  207.     If  a  line  be  drawn  through  a  point  in  the             ' 
bisector  of  an  angle  parallel  to  one  side  of  the  angle,  the 
bisector,  the  parallel,  and  the  other  side  of  the  angle  form 
an  isosceles  triangle.  qi^^ q 

Ex.  208.  If  the  median  to  the  base  of  a  triangle  is  perpendicular  to 
the  base,  the  triangle  is  isosceles. 

Ex.  209.  Prove  that  the  sum  of  the  perpendiculars 
drawn  from  any  point  in  the  base  of  an  isosceles  tri- 
angle to  the  equal  sides  of  the  triangle  is  equal  to  the 
altitude  drawn  to  one  of  the  equal  sides. 

Prove  0D+  0F=  CE. 

Suggestions.  —  1.    Draw  OG  1  CE. 

2.    Compare  OD  and  EG  ;  also  OF  and  CG. 

Ex.  210.  If  two  parallels  are  cut  by  a  transversal, 
the  bisectors  of  the  four  interior  angles  form  a  rec- 
tangle. 

Suggestions.  —  1.  EFGH  must  be  proved  a  O  and 
one  Z.  must  be  proved  a  right  angle. 

2.     Recall  §§  93,  103,  106. 

Ex.  211.  If  the  mid-point  of  any  side  of  a  square  is  joined  to  the  two 
vertices  of  the  opposite  side,  the  lines  so  drawn  are  equal. 

Ex.  212.  Prove  that  the  lines  drawn  from  the  mid-point  of  the  base  of 
an  isosceles  triangle  tcr  mid-points  of  the  sides  of  the  triangle  form  with  the 
half  sides  a  rhombus. 

Ex.  213.  If  the  lower  base  AD  of  trapezoid  ABCD  is  double  the  upper 
base  BG,  and  the  diagonals  intersect  at  E,  prove  that  GE'm  |  ^^and  that 
BE  \^  I  ED.     (§152.) 

Ex.  214.  If  D  is  any  point  in  side  ^(7  of  A  ABC  and  E,  F,  G,  and  H 
are  the  mid-points  of  AD,  CD,  BG,  and  AB,  respectively,  then  EFOH 
is  a  parallelogram. 

Suggestion.  —  Draw  BD. 

Note.  —  Supplementary  Exercises  86-108,  p.  281,  can  be  studied  now. 


BOOK   II 

THE   CIRCLE 

173.  Review  the  definitions  given  in  §  16  and  §  17,  and  the 
Exercises  31--34,  Introduction.  The  symbol  for  circle  is  O. 
The  circle  whose  center  is  0,  is  denoted  by  O  0. 

174.  Since  a  circle  is  a  closed  line  (§  6),  it  incloses  a  portion 
of  the  plane  called  its  interior. 

Ex.  1.  Draw  a  circle  with  radius  1  in.  Where  will  a  point  lie  :  («)  if 
its  distance  from  the  center  is  f  in.  ?  (6)  If  its  distance  from  the  center 
is  1.5  in.  '? 

Ex.  2.  Draw  a  circle  with  diameter  5  in.  Cut  the  circle  from  paper. 
Prove,  by  folding  it,  that  any  diameter  bisects  the  circle  and  the  surface 
within  the  circle. 

Ex.  3.  Draw  two  circles  which  intersect.  From  one  of  the  points  of 
intersection  draw  the  radius  of  each  circle.  How  does  the  distance  be- 
tween their  centers  compare  with  the  sum  of  their  radii  ? 

Ex.  4.    Prove  that  a  diameter  of  a  circle  is  greater 
than  any  other  chord  of  the  circle. 

Suggestion. — Compare  CD  with  AO  and  OB. 
Also,  compare  A0-\-  OB  with  AB. 

Ex.  5.     If  two  circles  intersect,  the  distance  between 
their  centers  is  greater  than  the  difference  of  their  radii. 

175.  From  the  exercises  following  §  17  and  §  174,  the  fol- 
lowing facts  are  evident : 

(a)  If  a  straight  line  cuts  a  circle,  it  intersects  it  in  two  and 
only  two  points. 

(b)  If  two  circles  intersect ,  they  have  two  and  only  two  points 
of  intersection. 

93 


94  PLANE    GEOMETRY  —  BOOK   II 

(c)  A  point  is  within,  on,  or  outside  a  circle  if  its  distance  from 
the  center  is  less  than,  equal  to,  or  greater  than  the  radius. 

(d)  A  diameter  of  a  circle  bisects  the  circle  and  the  surface  with- 
in it;  also,  if  a  line  bisects  a  circle j  it  is  a  diameter  of  the  circle. 

The  theorem  (d)  was  known  to  Thales. 

176.  One  half  of  a  circle  is  called  a  Semicircle. 
A  quarter  of  a  circle  is  called  a  Quadrant. 

Circles  having  the  same  center  are  called  Concentric  Circles. 

Peoposition  I.     Problem 

177.  Construct  a  circle  ivhich  loill  pass  through  three 
points  which  are  not  in  a  straight  line. 


Given  points  A,  B,  and  G  which  are  not  in  a  straight  line. 

Required  to  construct  a  circle  which  will  pass  through  A,  B, 
and  a 

Construction.     1.    Draw  AB  and  BC. 

2.    Construct  the  JL  bisectors  oi  AB  and  BC,  meeting  at  0. 

Statement.  A  circle  drawn  with  0  as  center  and  OA  as 
radius  will  pass  through  A,  B,  and  C. 

Proof.  OA=OB=  Oa  Why  ? 

Note,  —  Only  one  circle  can  be  drawn  through  three  points,  for  the 
center  must  lie  on  each  of  the  perpendicular-bisectors  (§  118,  II)  and 
these  lines  can  intersect  at  only  one  point. 

Ex.  6.  What  would  happen  if  the  three  points  in  Proposition  I  were 
in  a  straight  line  ? 


THE    CIRCLE 


95 


Ex.  7.     (a)  Construct  a  circle  which  will  pass  through   two  given 
points. 

(6)  How  many  circles  can  be  constructed  through  two 
given  points  ? 

(c)   Where  do  the  centerp  of  all  these  circles  lie  ? 

Eac.  8.     Construct  full  size  the  design  for  a  four-inch 
square  tile.     Make  the  decorative  arcs  f  in.  wide. 

178.  A  polygon  is  said  to  be  inscribed  in  a 
circle  when  its  vertices  lie  on  the  circle ;  as 
ABCD.  The  circle  is  said  to  be  circumscribed 
about  the  polygon. 

CHORDS,  ARCS,  AND  CENTRAL  ANGLES 

179.  Two  points  on  a  circle  are  the  ends  of 
two  arcs ;  a  Minor  Arc,  as  AMB,  and  a  Major  A 
Arc,  as  ANB. 

Unless  the  contrary  is  stated,  the  minor  arc  will 
always  be  understood  when  an  arc  is  indicated  by 
means  of  its  extremities.  Thus,  arc  AB  means  minor 
arc  AB. 

An  arc  AB  will  be  indicated  by  a  small  arc  drawn  over  AB  ;  as  AB. 

180.  A  Central  Angle  is  an  angle  whose  ver- 
tex is  at  the  center  and  whose  sides  are  radii 
of  the  circle ;  as  Z  ^OC 

Z  ^OC  is  said  to  intercept  AC. 

AC  is  said  to  be  iiitercepted  by  Z.  AOC, 

"  Intercept "  is  derived  from  two  Latin  words  meaning 


between 


and  "to  take,"  so  that  it  means  "  to  take  between." 

Eac.  9.  Construct  a  circle  with  radius  2  in.  (a)  Construct  two  cen- 
tral angles  which  are  equal,  and  a  third  central  angle  which  is  greater 
than  each  of  the  equal  central  angles.  (6)  Cut  from  the  paper  the  two 
equal  central  angles.     Compare  their  intercepted  arcs  by  superposition. 

(c)  Cut  from  the  paper  the  third  central  angle;  compare  its  inter- 
cepted arc  with  the  arc  intercepted  by  one  of  the  other  two  angles. 

(d)  What  do  you  conclude  must  be  true  about  the  arcs  intercepted  by 
equal  central  angles  of  a  circle  ?    By  unequal  central  angles  ? 


96  PLANE    GEOMETRY  —  BOOK   II 

Proposition  II.     Theorem 

181.  In  the  same  circle  or  in  equal  circles^  if  central 
angles  are  equal,  they  intercept  equal  arcs. 


Hypothesis.  ©  0  =  O  i? ;   Z  AOB  =  Z  CRD. 

Conclusion.  AB  ^  CD. 

Proof.     1.   Place  O  0  on  Q  R,  with  point  0  on  point  R, 
and  so  that  Z  AOB  coincides  with  its  equal,  Z  CRD. 

2.  Then  O  0  will  coincide  with  Q)  R.  §  17 

3.  Point  A  will  fall  on  point  C,  since  OA  =  RC        §  17 

4.  Point  B  will  fall  on  point  D,  since  OB  =  RD.        §  17 

5.  .-.  AB  coincides  with  CD  and  hence  AB  =  CD. 

Ex.  10.     Divide  a  circle  into  four  equal  arcs. 
What  kind  of  central  angles  must  be  constructed  ? 

Ex.  11.     Divide  a  circle  into  eight  equal  arcs. 
Ex.  12.     Divide  a  circle  into -six  equal  arcs. 

How  large  must  the  central  angles  be  ?    Recall  §  70  and  Ex.  113,  Book  I. 

Ex.  13.     Using    the    construction    made    in   Ex.    12,       /^TV^ 

draw  a  six-pointed  star.  A"/ '  ^/\ 

Ex.  14.     Tell  how  you  can  divide  a  circle  into  five    \/\       /\j 
equal  parts   by  means  of  your  protractor  and  straight-     \/\'7'y 
edge. 

Ex.  15.  By  means  of  compass,  ruler,  and  protractor,  L- — /  \ — A 
draw  a  five-pointed  star  in  a  circle  with  2  in.  radius,  to  I  ^  ^  j 
be  used  as  a  pattern  for  a  star  on  a  sailor  collar.  \  //\\/ 


THE    CIRCLE 


97 


Proposition  III.     Theorem 

182.  In  the  sam,e  circle  or  in  equal  circles,  if  arcs  are 
equal,  the  central  angles  ivhich  intercept  them  are  equaL 


Hypothesis.  Q  0=  Q  B,  AB=  CD. 

Conclusion.  Z  AOB  =  Z  CRD. 

Proof.  1.  Since  O  0  =  O  i2,  and  J2  =  CD,  the  O  0  can 
be  made  to  coincide  with  the  O  Rj  and  AB  with  CD,  A 
falling  on  C,  B  on  D,  and  0  on  R. 

2.  .-.  AO  will  fall  on  CR  and  BO  on  DR.       Ax.  10,  §  15 

3.  .-.  Z  AOB=  Z  CRD.  Why  ? 


Ex.  16.  If  a  radius  bisects  an  arc,  it  is  perpendicular 
to  and  bisects  the  chord  which  subtends  the  arc. 

Ex.  17.  The  twelve  spokes  of  a  wheel  are  spaced 
so  that  the  points  at  which  they  are  attached  to  the  rim 
divide  the  rim  into  equal  arcs.  How  many  degrees  are 
there  in  the  angle  formed  by  two  adjacent  spokes  ? 


183.  It  may  be  proved  that :  in,  the  same  circle  or  in  eqxtal 
circles, 

(a)  The  greater  of  two  unequal  central  angles  intercejits  thi 
greater  arc; 

(b)  The  greater  of  two   unequal  arcs   is   intercepted   by  the 
greater  central  angle. 

184.  A  chord  AB  is  said  to  subtend  arc  AB.     Arc  AB  is 
said  to  be  subtended  by  chord  AB, 

*'  Subtend  "  is  derived  from  Latin  words  meaning  "  to  stretch  under." 


98 


PLANE    GEOMETRY  —  BOOK   II 


Pkopositiois'  ly.     Theorem 

185.    In  the  same  circle  or  in  equal  circles,  if  chords 
are  equal,  they  subtend  equal  arcs. 


Hypothesis.  O  0  =  Q  R)  AB=  CD. 

Conclusion.  AB  =  CD. 

Plan.     1.  Draw  AO,  OB,  RC,  RD. 

2.  Prove  Z  0  =  Z  i?,  and  apply  §  181. 

[Proof  to  be  given  by  the  pupil.] 

Ex.  18.  Construct  an  equilateral  triangle.  Circumscribe  a  circle 
about  the  triangle.  (§  177.)  Prove  that  the  vertices  of  the  triangle  di- 
vide the  circle  into  three  equal  arcs. 


Proposition  V.     Theorem 

186.    In  the  same  circle  or  in  equal  circles,  if  arcs 
are  equal,  the  chords  which  subtend  them  are  equal. 

Hypothesis.  QO  =  Q  R;  AB=CD.     (Fig.  §  185.) 

Conclusion.  AB  =  CD. 

Plan.    Try  to  prove  AAOB^A  CRD.     Compare  Z  0  and 

Z  R.  §  182 

[Proof  to  be  given  by  the  pupil.] 

Q 

Ex.  19.     If  C  is  the  mid-point  of  arc  AB,  prove  that 
AC  is  greater  than  one  half  AB.  j^ 

Draw  CB.     Compare  AB  with  AC  +  CB. 


THE    CIRCLE 


99 


Proposition  VI.     Theorem 

187.  In  the  same  circle  or  in  equal  circles,  if  two 
minor  arcs  are  unequal,  then  their  chords  are  unequal, 
the  greater  arc  heing  subtended  by  the  greater  chord. 


Hypothesis.  O  0  =  O  /2 ;  AB>  CD. 

Conclusion.  AB  >  CD. 

Proof.     1.       Draw  radii  AO,  BO,  CR,  and  DR. 

2.  In  A^O^and  AC/eZ>: 
AO=CRsind  BO  =  DR', 

but  since  AB>  CD,  Z  0>  ZR. 

3.  .-.  AB  >  CD. 


Hyp. 

§  183,  b 
§166 


Proposition  VII.     Theorem 

188.  In  the  same  circle  or  in  equal  circles,  if  two 
chords  are  unequal,  then  they  subtend  unequal  minor 
arcs,  the  greater  chord  subtending  the  greater  arc. 

Hypothesis.  Q  0  =  Q  R;  AB>  CD.     (Fig.  §  187.) 

Conclusion.  AB  >  CD. 

Plan.     Prove  ZO>ZR(%  167)  and  apply  §  .183,  a. 

D 

Ex.  20.  Prove  that  the  straight  line  which  bisects 
the  arcs  subtended  by  a  chord  bisects  the  chord  at  right 
angles. 

Suggestion.  —CompaiTe  AD  and  BD;  also  AC  and  BC. 
Apply  §  77. 


100 


PLANE    GEOMETRY  —  BOOK   II 


Pkoposition  Yin.     Theorem 

189.    If  a  diameter  is  perpendicular  to  a  chord,  it 
bisects  the  chord  and  its  subtended  arcs. 


D 

/i  ix 

V  /      3  4 X 

^K — Y~~y 


Hypothesis.  In  O  0,  diameter  CD  ±  AB  at  E. 

Conclusion.  AE  =  EB;  AC=CB',  smd  AD  =  DB. 

Proof.     1.  Draw  AO  and  OB. 

2.  A  AEO  ^  A  OEB.      Give  the  full  proof. 

3.  .'.  AE  =  EB,  and  Z.1  =  A2.  Why? 

4.  ,\AC=GB.  §181 

5.  Also,  AD  =  DB.                                   Why? 

190.  Cor.  1.     A  line  through  the   center  of  a   circle  perpen- 
dicular to  a  chord  bisects  the  chord. 

191.  Cor.  2.  The  p>erpendicular-bisector  of  a 
chord  passes  through  the  center  of  the  circle,  and 
bisects  the  arcs  subtended  by  the  chord.  A 

Compare  AO  and  OB.     Then  use  §  118,  II 

Ex.  21.     If  a  radius  of  a  circle  bisects  a  chord,  it  is  perpendicular  to 
the  chord  and  bisects  the  subtended  arcs. 

Ex.  22.     Determine  the  center  and  the  radius  of  the  circle  of  which 
AB  is  an  arc. 

Draw  any  two  chords  and 
erect  the  ±  bisectors.  These 
must  pass  through  the  center. 
(§  191.)  A  B 


THE   giRCLE  101 

Proposition^' 'IX..    Xu^o,rsm, 

192.   In  the  same  circle' or  in  equal  circles ,  if  chords 
are  equal,  they  are  equidistant  from  the  center. 


Hypothesis.  In  OABC: 

AB=GD;  OE±AB',   OF  A.  CD. 
Conclusion.  OE  =  OF. 

Proof.     1.  Draw  OA  and  0(7. 

2.  AE  =  \AB,CF^\CD,zxiiiAB=CD.      Why? 

3.  .'.AE=CF.  Ax.  6,  §51 

4.  .-.  A  AEO  ^  A  CFO.  Give  full  proof. 
6.                                 .'.OE=OF.  Why? 


Ex.  23.  If  two  intersecting  chords  are  equal,  the  radius  drawn 
through  the  point  of  intersection  bisects  the  angle  between  them. 

Ex.  24.  If  a  straight  line  bisects  a  chord  and  its  subtended  arc, 
then  it  is  perpendicular  to  the  chord.     (§  186. ) 

Ex.  25.  If  a  straight  line  is  drawn  cutting  two  concentric  circles  in 
the  points  A,  B^  C,  and  D,  respectively,  then  AB  equals  CD. 

Ex.  26.  Prove  that  the  perpendicular-bisectors  of  the  sides  of  an 
inscribed  polygon  are  concurrent.     (§  168.) 

Ex.  27.  On  equal  chords  of  a  circle,  points  are  taken  at  equal 
distances  from  the  ends  of  the  chords.  Prove  that  all  these  points  are 
equidistant  from  the  center  of  the  circle. 


102  PLANE    GEOMETRY  —  BOOK   II 

Propqsitiok  .  X.     Theorem 

193.    In  the  same  circle  or  in  equal  circles,  if  chords 
are  equidistant  from  the  center,  they  are  equal. 

^^ — "^^^ 


Hypothesis. 

In  O^BC: 
OE^^AB;  0F1.CD)  OE=OF. 

Conclusion. 

AB  =  CD. 

Proof.     1. 

Draw  OA  and  OC. 

2.     Then 

A  OEA^AOCF  3ind  AE  =  OF. 
[Give  the  full  proof.] 

3.     But 

AB  =  2AE  and  CD  =  2  CF. 

Why? 

4. 

.-.  AB  =  CD. 

Why? 

tJx.  28.  If  a  straight  line  be  drawn  parallel  to  the  line  connecting 
the  centers  of  two  equal  circles,  and  intersecting  the  circles,  then  the 
chords  formed  in  the  two  circles  are  equal. 

Ex.  29.  If  two  chords,  intersecting  within  the  circle,  make  equal 
angles  with  the  radius  passing  through  their  point  of  intersection,  they 
are  equal. 

Suggestion.  — DrsiW  Js  to  the  chords  from  the  center;  prove  the  Js  are 
equal ;  then  use  §  193. 

Ex.  30.  If  two  equal  chords  of  a  circle  intersect,  the  segments  of  one 
equal  respectively  the  segments  of  the  other. 

194.  A  straight  line  which  intersects  a  circle  in  two  points 
is  called  a  Secant  of  the  circle. 

Ex.  31.  If  ABE  and  DCE  are  two  secants 
of  a  circle  which  make  equal  angles  with  the 
line  connecting  E  with  the  center  of  the  circle, 
then  chord  AB  =  chord  DC.     (Use  §  193.) 


THE    CIRCLE  103 

Proposition  XI.     Theorem 

195.  In  the  same  circle  or  in  equal  circles^  the  less  of 
tivo  unequal  chords  is  at  the  greater  distance  from  the 
center  of  the  circle. 


E 

Hypothesis.  In  O  0 : 

AB<CD;  OF±AB',  OG  ±  OD. 

Conclusion.  OF  >  OG. 

Proof.     1.        Since  AB  <  CD,  AB  <  CD.  §  188 

2.  Let  CE  =  AB.     Draw  CE. 

3.  .\CE  =  AB.  Why? 

4.  Draw  OH  A,  CE,  intersecting  CD  at  K. 

5.  .-.  011=  OF.  §  192 

6.  But  OH  >  OK.  Why  ? 

7.  .\OF>OK.  Why? 

8.  OK>OG.  Why? 

9.  .\OF>OG.  Why? 

Ex.  32.  All  equilateral  tiiani^'le  and  a  square  are  inscribed  in  a  circle. 
Prove  that  the  sides  of  the  triangle  are  nearer  the  center  than  the  sides  of 
the  square. 

Ex.  33.  Chord  BY  is  drawn  through  one  extremity  of  a  diameter 
AB  of  circle  0.  Radius  OX  is  drawn  in  Z^BF  parallel  to  BY,  inter- 
secting arc  ^  F  at  X     Prove  arc  ^X  equals  arc  XF. 

Suggestion. — Draw  radius  OY. 

Ex.  34.  AB  is  a  diameter  of  a  circle  and  XF  is  an  intersecting 
diameter  of  a  smaller  concentric  circle.     Prove  AXB  Y  is  a  parallelogram. 


104 


PLANE    GEOMETRY  —  BOOK   II 


Proposition  XII.     Theorem 

196.  In  the  same  circle  or  in  equal  circles,  if  tivo 
chords  are  unequally  distant  from  the  center,  the  more 
remote  is  the  smaller. 


Hypothesis.  In  O  O : 

OELAB',  OF±CD,  0E>  OF. 
Conclusion.  AB  <  CD. 

Proof.     1.  Suppose  that  AB  is  not  less  than  CD ;  that  is, 
suppose  that  AB  =  CD  or  AB  >  CD. 

2.  If  AB  =  CD,  then  OE  =  OF. 

But  OE  >  OF. 
.'.AB  is  not  =  to  CD. 

3.  liAB>  CD,  then  OE  <  OF. 
But  OE  >  OF. 

.'.  AB  is  not  greater  than  CD. 

4.  ..AB<CD. 


Why? 
Why? 

§195 
Hyp. 


197.  Tangent  Line.  Assume  that  the  secant  AB  turns  about 
the  point  A  in  the  direction  indicated  by 
the  arrow.  The  point  B  moves  closer  to 
the  point  A.  When  B  finally  coincides 
with  A,  the  line  assumes  the  position  XY. 
XYis  called  a  tangent  to  the  circle. 

A  tangent  to  a  circle  is  a  straight  line 
which  touches  the  circle  at  only  one  point. 

The  circle  is  also  said  to  be  tangent  to  the  line. 

The  point  where  the  tangent  touches  the  circle  is  called  the 
Point  of  Tangency,  or  Point  of  Contact. 


THE   CIRCLE 


105 


Pkoposition  XIII.     Theorem 

198.  A  straight  line  perpendicular  to  a  radius  at  its 
outer  extremity  is  a  tangent  to  the  circle. 


Hypothesis. 

Conclusion. 

Proof.    1. 

2. 

3. 

4. 


OC  is  a  radius  of  O  0 ;  ^J5  ±  OC  at  G. 


AB  is  tangent  to  O  0. 

Let  D  be  any  point  in  AB  except  O. 
Draw  OD. 
.'.  OD  >  00.  Why  ? 

point  D  lies  outside  of  the  ©.  §  175,  c 

5.     .*.  every  point  in  AB  except  0  lies  outside  the  circle, 
and  hence  AB  is  tangent  to  the  circle.  §  197 


199.  Cor.  1.  A  tangent  to  a  circle  is 
perpendicular  to  the  radius  drawn  to  the 
point  of  contact. 

Since  all  points  in  AB  except  C  lie  outside  the 
circle,  OC  Is  the  shortest  segment  to  AB  from  0. 
Hence  OCXAB. 

200.  Cor.  2.  A  line  perpendicular  to  a 
tangent  at  its  point  of  contact  passes  through 
the  center  of  the  circle. 

By  Cor.  1,  the  radius  00  is  ±  to  AB.    Hence 


OC  and   CD  must  coincide.     (Why?)    .• 
passes  through  the  center  of  the  circle. 


CD 


201.   Cor.  3.     A  line  from   the  center  of  the  circle  perpen- 
dicular to  a  tangent  passes  through  the  point  of  contact. 


106 


PLANE    GEOMETRY  —  BOOK   II 


Proposition  XIV.     Theorem 

202.   The  tangents  to  a  circle  from  an  outside  point 
are  equal. 


Hypothesis.     AB  and  AC  are  tangents  to  O  0. 
Conclusion.  AB  =  AC. 

[Proof  to  be  given  by  the  pupil.     Recall  §  114.] 

Note.  —  The  proof  of  Prop.  XIV  is  attributed  to  a  mathematician  Fink, 
with  the  date  1583.  The  theorem  does  not  appear  in  Euclid  at  all.  It 
appears  first  as  a  definite  theorem  in  writings  of  Hero,  although  it  was 
apparently  used  by  Archimedes. 

Ex.  35.  Prove  that  the  tangents  to  a  circle  at  the  extremities  of  a 
diameter  are  parallel.  ^ 

Ex.  36.     If  two  circles  are  concentric,  any  two  chords 
of   the  greater  which  are  tangents  of  the  smaller  are    C 
equal. 


Ex.  37.  Prove  that  all  tangents  drawn  from  the  larger  of  two  con- 
centric circles  to  the  smaller  are  equal. 

Ex.  38.  Prove  that  the  line  joining  the  center  of  a  circle  to  the  point 
of  intersection  of  two  tangents  :  (a)  bisects  the  angle  formed  by  the 
radii  drawn  to  the  points  of  contact ;  (6)  bisects  the  angle  formed  by  the 
tangents  ;  (c)  bisects  and  is  perpendicular  to  the  chord  joining  the  points 
of  contact.  c 

Ex.  39.  Prove  that  the  sum  of  two  opposite  sides  of  ^ 
a  circumscribed  quadrilateral  is  equal  to  the  sum  of  the  e 
other  two  opposite  sides. 


THE   CIRCLE 


107 


203.   A  straight  line  tangent  to  each  of  two  circles  is  called 
a  Common  Tangent  of  the  circles; 
as  AB. 

If  the  circles  lie  on  opposite  sides 
of  AB,  AB  is  called  a  common  in- 
ternal tangent. 

If  the  circles  lie  on  the  same  side 
of  AB,  AB  is  called  a  common  ex- 
ternal tangent. 

The  length  of  a  common  tangent 
is  the  length  of  the  segment  between 
the  two  points  of  contact. 

Some  uses  of  common  tangents 
are  pictured  in  the  figures  below. 


Belts  abound  Pulleys 


Chain  around  Wheels 


Ex.  40.  Prove  that  the  coniraon  internal  tangents  of  two  circles  are 
equal. 

Ex.  41.  Trove  that  the  common  external  tangents  of  two  circles  are 
equal,  when  the  circles  are  unequal. 

Note.  —  The  theorem  is  also  true  when  the  circles  are  equal.  This 
might  be  solved  as  an  optional  exercise. 


204.  Two  circles  are  tangent  when 
they  are  tangent  to  the  same  straight 
line  at  the  same  point. 

They  are  tangent  externally  if  they 
lie  on  opposite  sides  of  the  common 
tangent. 

They  are  tangent  internally  if  they 
lie  on  the  same  side  of  the  tangent. 

Note.  —  Supplementary  Exercises  1-6,  p. 
283,  can  be  studied  now. 


108  PLANE    GEOMETRY  —  BOOK   II 

Proposition  XV.     Theorem 

205.   If  two  circles  are  tangent  to  each  other,  their 
line  of  ceyiters  passes  through  the  point  of  co7itact. 


Hypothesis.     (D  0  and  0'  are  both  tangent  to  AB  at  A. 

00'  is  the  line  of  centers. 
Conclusion.         St.  line  00'  passes  through  A. 
Proof.    1.         Draw  the  radii  OA  and  O'A. 

2.  Then  OA  J_  AB  and  also  O'A  _L  AB.  Why  ? 

3.  .-.  OAO'  is  a  st.  line.  §  40 

4.  .-.  St.  lines  00'  and  OAO'  coincide.         Ax.  10,  §  51 

5.  .-.  00'  passes  through  A. 

Note. — The  theorem  has  been  proved  for  two  ©  tangent  externally. 
As  an  optional  exercise,  it  is  suggested  that  the  theorem  be  proved  when 
the  (D  are  tangent  internally. 

206.  Cor.  If  the  distance  between  the  centers  of  two  circles 
equals  the  sum  of  their  radii,  the  circles  are  tangent  externally. 

For  then  a  point  A  can  be  taken  on  00'  so  that  OA  =  one  radius  and 
then  O'A  =  the  other  radius.  A  perpendicular  to  00'  at  J.  will  then  be 
tangent  to  each  of  the  circles.  Hence  the  circles  are 
tangent  (§  204). 

Ex.  42.  Study  the  adjoining  figure  to  determine  how 
to  construct  it.  Construct  a  figure  like  it,  having  the  radius 
of  the  large  circle  1  in.  and  that  of  the  small  circles  |  in. 

Ex.  43.     How  many  common  tangents  do  two  circles  have 
(a)  Which  are  tangent  internally  ? 
(&)  Which  are  tangent  externally  ? 


THE    CIRCLE 


109 


Proposition  XVI.     Theorem 

207.  If  tivo  circles  intersect,  the  straight  line  joining 
their  centers  bisects  their  common  chord  at  right  angles. 


Hypothesis.     (D  0  and  0'  intersect  at  A  and  B. 
AB  is  the  common  chord  and  00'  is  the  line  of  centers. 

Conclusion.     00'  A.  AB  and  00'  bisects  AB. 

Suggestion.  — Btslw  6 A,  OB,  O'A,  and  O'B.     (Apply  §  77.) 

Ex.  44.  If  two  circles  0  and  0'  intersect  at  points  A  and  B^  and  if 
00'  intersects  O  O  at  Xand  O  0'  at  F,  then  Xand  Fare  each  equidis- 
tant from  A  and  B. 

Ex.  45.  If  a  straight  line  be  drawn  through 
the  point  of  contact  of  two  circles  which  are 
tangent  externally,  terminating  in  the  circles, 
the  radii  drawn  to  its  extremities  are  parallel. 

Note.  — The  theorem  is  stated  for  two  circles  which  are  tangent  exter- 
nally.   Investigate  its  truth  for  two  circles  which  are  tangent  internally. 

Ex.  46.  If  two  circles  are  tangent  to  each  other  externally  at  point 
A,  the  tangents  to  them  from  any  point  in  their  common  tangent  which 
passes  through  A  are  equal. 


Ex.  47.  If  two  circles  are  tangent  to  each  other 
externally  at  point  A,  the  common  tangent  which  passes 
through  A  bisects  the  other  two  common  tangents. 


Ex.  48.  AB  and  AC  are  the  tangents  to  a  circle  from  point  A^  and 
D  is  any  point  in  the  smaller  of  the  arcs  subtended  by  the  chord  BC.  If 
a  tangent  to  the  circle  at  D  meets  AB  at  F,  and  AC  at  F,  prove  the  per- 
imeter of  A  AFF  =  AB  +  A'^. 


no 


PLANE    GEOMETRY  —  BOOK   II 


Proposition  XYII.     Theorem 
208.  Parallel  lines  intercept  equal  arcs  on  a  circle. 
Case  I.      When  one  line  is  a  tangent  and  one  a  secant: 

E 


Hypothesis. 
Conclusion. 

AB 

is  tangent  to  O  CED  at  E ; 
secant  CD  li  AB. 

CE  =  DE.' 

Proof.   1. 

2. 

3. 

4. 

Draw  diameter  EF. 
..EF±AB. 
.'.  EF±  CD. 
.:  CE  =  DE. 

Why? 
Why? 

§  189 

Case  II.     When  both  lines  are  secants . 

Hypothesis.     AB  and  CD  are  II  secants 

of  O  ABDC. 
Conclusion.  AC  =  BD. 

Proof.  1.    Assume  J5;6^i^ tangent  to  the 
O  at  6r,  and  parallel  to  CD. 


E  O  F 

(Complete  the  proof.     Compare  AG  and  BG  ;  also  CG  and  D^. ) 

E 


Case  III.      When  both  lines  are  tan- 
gent:  q 

Hypothesis.     AEB    and    CFD    are 

1!  tangents  to  the  O  at  JS'  and  F. 

Conclusion.  EGF=  EHF\ 

[Proof  to  be  given  by  the  pupil.]  C" 


THE   CIRCLE 


111 


Ex.  49.     Prove  that  the  straight  line  joining  the  points  of  contact 
of  two  parallel  tangents  of  a  circle  is  a  diameter  of  the  circle. 
Ex.  50.     Prove  that  an  inscribed  trapezoid  must  be  isosceles. 

Q 

Ex.  51.     The  adjoining  figure  gives  the  method  of        b^^^^^^^^'^^^^^^d 
construction  of  one  form    of    mansard  roof.     The 
chords  AB,  BC,  CD,  and  DE  are  equal. 

(a)  Construct  such  a  figure  for  a  roof  whose  span  AE  is  28',  using  the 
scale  J"  =  1'. 

(6)  Is  the  line  BD  parallel  to  AE  ?    Prove  it. 

209.  Theorems  concerning  tangents  and  tangent  circles  have 
unusually  wide  application  in  design. 

Direction  is  naturally  indicated  by  a  straight  line. 

On  a  circle,  the  direction  is  constantly  changing.  It  is  con- 
venient in  both  pure  and  applied  mathematics 
to  speak  of  the  direction  of  a  curve  at  a  point ; 
also  it  is  agreed  that  this  direction  shall  be  the 
same  as  the  direction  of  the  tangent  to  the 
curve  at  the  point.  It  is  this  fact  which  is 
used  in  a  variety  of  ways. 


If  a  road  turns  a  corner  as  pictured,  there  is  an 
abrupt  change  of  direction.  If  a  street  car  line  runs 
along  the  road,  such  an  abrupt  change  in  direction  in 
the  tracks  is  impossible.  For  that  reason,  the  arrange- 
ment of    tracks   indicated   in  the  adjoining     

figure  is  employed.     A  car  running  from  C     ^_^ 

towards  D  passes  readily  from  CA  to  the  arc    ^_ 

AB,  for  on  both  the  straight  line  and  the  arc 
the  direction  at  A  is  the  direction  of  line 
CA  ;  similarly  at  B. 

Ex.  52.  Where  must  the  center  0  be  lo- 
cated in  order  that  the  circle  will  be  tangent  to 
both  CA  and  BD  in  the  last  figure  for  §  209  ? 

Ex.  53.  What  kind  of  circles  should 
circles  AB  and  EF  be  ? 

Ex.  54.  Determine  how  to  construct 
the  figures  at  the  right.  Construct  such 
figures  in  circles  with  diameter  3  in. 


~7|  B 
-'O 


112  PLANE    GEOMETRY  —  BOOK   II 

MEASUREMENT   OF  ANGLES  AND  ARCS 

210.  To  measure  a  given  magnitude,  two  steps  are  necessary. 

(a)  Select  a  quantity  of  the  same  kind  to  be  used  as  the  unit 
of  measure. 

(b)  Determine  the  number  of  times  the  given  magnitude  con- 
tains the  unit  of  measure.  This  number  is  called  the  Numerical 
Measure  of  the  quantity  in  terms  of  the  unit  employed. 

If  the  quantity  contains  the  unit  itself  or  any  part  of  it  an 
integral  number  of  times,  the  quantity  can  be  measured  exactly. 

If  the  quantity  does  not  contain  the  unit  of  measure  an  in- 
tegral number  of  times,  the  quantity  can  be  measured  only 
approxi7nateli/. 

Thus,  the  diagonal  of  a  square  whose  side  is  1  in.  is  known  to  be 
1.414  +  in.,  where  the  decimal  is  a  "never  ending "  decimal. 

211.  Two  magnitudes  of  the  same  kind  are  said  to  be  Com- 
mensurable when  each  contains  the  same  unit  of  measure,  called 
a  Common  Measure,  an  integral  number  of  times. 

Thus,  two  segments  whose  lengths  are  2|  in.  and  3\  in.  respectively 
are  commensurable,  for  the  common  measure  ^  in,  is  contained  in  the  first 
segment  10  times  and  in  the  second  13  times. 

Two  magnitudes  of  the  same  kind  are  said  to  be  Incommensur- 
able when  no  unit  of  measure  can  be  found  which  is.  contained 
an  integral  number  of  times  in  each. 

The  diagonal  and  the' side  of  a  square  are  incommensurable. 

212.  The  Ratio  of  two  magnitudes  of  the  same  kind  is  the 
quotient  of  their  numerical  measures  in  terms  of  a  common 
measure. 

Thus,  the  segments  of  lengths  2^  in.  and  3|  in.,  in  §  211,  have  the 
ratio  \^  . 

Ex.  55.  What  is  the  measure  of  a  yard  in  terms  of  the  unit :  {a)  1  ft.  ? 
(6)  1  in.  ?  (c)  \  in.  ? 

Ex.  56.  What  is  the  measure  of  1  gallon  in  terms  of  the  unit :  (a)  1 
qt.?     (ft)  1  pt.?     (c)  1  gill? 

Ex.  57.     What  is  the  ratio  of  2  yd.  to  1^  ft.? 


MEASUREMENT   OF  ANGLES  AND  ARCS         113 

Proposition  XVIII.     Theorem 

213.   In  the  same  circle  or  in  equal  circles,  two  central 
angles  have  the  same  ratio  as  their  intercepted  arcs. 


Case  I.      When  the  angles  are  commensurable : 
Hypothesis.     In  O  0,  Z  AOB  and  Z  BOC  are  commensur- 
able. 
ConclusioD.  ^QB^AB, 

ZBOC     ^ 

Proof.     1.      Z  AOB  and  Z  BOC  have  a  common  measure. 

§211 

2.  Let  the  common  measure  be  Z  AOD,  and  let  it  be  con- 
tained in  Z.AOB  4  times  and  in  Z  BOC  3  times. 

3.  ...^^^  =  1  2^2 

ZBOC     3  _ 

4.  The  radii  drawn  from  0  in  step  (2)  divide  AB  into  4  and 
BC  into  3  arcs  which  are  all  equal.  Why  ? 

6.  .-.^  =  1  §212 

6.     .*.  from  steps  (3)  and  (5), 

Z  BOC      BC 

Case  II.      TF^e^i  the  angles  are  incommensurable : 

The  theorem  is  true  also  in  this  case.     The  proof  presents 

certain  difficulties  which  it  is  wise  to  postpone  at  this  time. 

This  proof  is  taken  up  in  §  423. 


114  PLANE   GEOMETRY  —  BOOK  II 

Ex.  58.  In  the  adjoining  figure,  compare  AB  and 
BC.    Also  compare  AB  and  DC;  also  BC  and  DC. 

Ex.  59.  A  right  central  angle  is  what  part  of  a 
straight  angle  ?  What  part  therefore  is  its  intercepted 
arc  of  a  semicircle  ? 

Ex.  60.  A  60°  angle  is  what  part  of  the  perigon  ? 
What  part  therefore  is  its  intercepted  arc  of  the  whole 
circle  ? 

Ex.  61.  If  a  circle  is  divided  into  5  equal  parts,  what  part  of  the 
perigon  is  the  central  angle  which  intercepts  one  of  the  parts?  How 
many  degrees  are  there  in  the  central  angle  ? 

Ex.  62.  If  AB^  any  chord  of  circle  O,  is  extended  to  a  point  C  so  that 
BC  equals  the  radius  of  the  circle,  and  CO  is  drawn,  cutting  the  circle  at 
Z  and  E  respectively,  then  AE  =  3  •  BZ. 

Suggestions.  — 1.  Draw  OA  and  OB.  2.  FroYe  A  AOE  =  3  jLBOC.  Use 
§  87  and  §  69. 

214.  Measuring  Angles  and  Arcs.  In  §  28,  the  unit  for 
measuring  angles  is  given  as  1  degree,  J^-  of 
a  right  angle  or  g-^-g-  of  the  perigon.  This  will 
be  called  for  the  present  one  angular-degree. 
Let  Z  AOB  represent  1°.  Similarly,  we  shall 
speak  of  angular-minutes  and  angular-seconds. 
Thus,  60  angular-seconds  equal  one  angular- 
minute  ;  and  60  angular-minutes  equal  one  angular-degree. 

Let  a  circle  be  drawn  around  point  0  as  center,  and  the 
radii  which  divide  the  perigon  into  360  equal  central  angles 
be  imagined.  These  angles  are  angular-degrees.  They  will 
intercept  360  equal  arcs  on  the  circle.  Let  AB  represent  one 
of  these  arcs.  It  is  the  unit  for  measuring  arcs  on  this  circle 
and  on  any  equal  circle.     It  will  be  called  one  arc-degree. 

Evidently  on  a  circle  with  longer  radius,  the  arc  corresponding  to  AB 
will  be  longer. 

In  similar  manner,  each  arc-degree  could  be  divided  into  60 
equal  parts,  called  arc-minutes,  and  each  arc-minute  into  60 
equal  arc-seconds.  A  central  angle  of  one  angular  minute 
intercepts  an  arc  of  one  arc-minute. 


MEASUREMENT   OF  ANGLES  AND  ARCS        115 

215.  A  central  angle  has  the  same  measure  as  its  inter- 
cepted arc,  ivhen  angular-degrees  and  arc  degrees  are  used  as  the 
respect  ice  units  of  measure. 

Let  Z  AOB  represent  1  angular-degree  and  AAOC  any- 
other  central  angle. 

Then  ZAOC^AC  ^  213 

ZAOB     ^  ^ 

Z.  AOO  AC 

But  —    is  the  numerical  measure  Z  AOC,  and  — —   is 

ZAOB  ^  ^ 

the  numerical  measure  of  AC,  by  the  definition. 

Hence  the  measure  of  Z  AOC  equals  the  measure  of  AC. 

Thus,  if  Z  AOC  =  57.29  angular-degrees,  then  AV=  57.29  arc- 
degrees. 

From  now  on,  it  will  be  understood  that  angles  are  measured 
in  terms  of  angular-degrees,  and  arcs  in  terms  of  arc-degrees. 
Also,  the  following  statement  of  the  theorem  of  §  215  will  be 
employed  for  convenience : 

A  central  angle  is  measured  by  its  intercepted  arc. 

Ex.  63.     What  is  an  arc-degree  ?     An  angular- degree  ? 

Ex.  64.     Are  all  angular-degrees  of  the  same  size  ? 
Are  all  arc-degrees  of  the  same  size  :   (a)  on  the  same  or  on  equal 
circles  ?  (6)  on  Unequal  circles  ? 

Ex.  65.  If  ABCD  is  an  inscribed  square  and  0  is  the  center  of  the 
circle,  how  may  degrees  are  there  in  AB  ?  in  Z  AOB  ? 

Ex.  66.  A  ABC  is  an  equilateral  triangle  inscribed  in  a  circle  with 
center  O  ;    how  many  degrees  are  there  in  AB  ?  in  ZAOB  ? 

216.  An  angle  is  said  to  be  an  Inscribed  Angle  when  its 
vertex  is  on  the  circle  and  its  sides  are  chords 
of  the  circle  ;  as  Z  ADC. 

Z ABC  intercepts  the    AC;    ^C  is   intercepted 
hyWxQZABC. 

Such  an  angle  is  said  to  be  inscribed  in  a  g 

circle  or  may  be  said  to  be  inscribed  in  the  arc  ABC 


116 


PLANE    GEOMETRY  —  BOOK   II 


Proposition  XTX.     Theorem 

217.    An  inscribed  angle  is  measured  hy  one  half  its 
intercepted  arc. 

Case  I.      Wlieu  07ie  side  of  the  angle  is  a  diameter : 

A 


Hypothesis. 


Conclusion. 

Proof.     1. 

2. 

3. 

4. 

5. 

6. 


i^l. 


Why? 
Why? 

§215 


AC  is  a  diameter ;  AB  is  any  other  chord  of 

OO.  _ 

Z  BAC  is  measured  by  i  BC. 
Draw  BO. 
Z1  =  ZB+ZA, 
ZB  =  Z  A. 
.-.  Z  1  =  2  .  Z  ^,  or  Z  ^  = 

But  Z  1  is  measured  by  BC. 
.-.  Z  A  is  measured  by  ^  BG. 

Case  II.      When  the  center  of  the  Q   is 
within  the  angle : 

Hypothesis.     Center   0  lies   within 

.     scribed  Z  BAC. 
Conclusion.       Z.BAC   is    measured 

iBC. 
Proof.     1.     Draw  diameter  AD. 

2.  Then  Z  1  is  measured  by  |  BD, 
and  Z  2  is  measured  by  i  DC  Case  I 

3.  .-.  Z  1  +  Z  2  is  measured  by  |  (BD  +  DC),    Ax.  3,  §  51 
or  Z  BAC  is  measured  by  |  BC, 


MEASUREMENT   OF  ANGLES  AND  ARCS 


117 


Case  III.     When  the  center  of  the  O  lies 
outside  the  angle : 

Hypothesis.     Center   0  lies   outside  in- 
scribed Z  BAC. 
Conclusion.     Z  BAC    is    measured    by 
\BC. 

Suggestions.  —  1.    Z  BAD  is  measured  by  what  ? 
2.    LCAD'i 

218.   Cor.  1.     An    angle    inscribed   in   a 
semicircle  is  a  right  angle. 

(If  BCis  a  diameter,  prove  Z  A  =  \ri.  A.) 


Inscribed  angles  which  intercept  the  sayne  arc 


219.   Cor.  2. 

are  equal. 

Ex.  67.  Three  consecutive  sides  of  an  inscribed  quadrilateral  sub- 
tend arcs  of  82°,  90°,  and  60°  respectively.  Find  each  angle  of  the  quad- 
rilateral. 

Ex.  68.  Construct  a  line  perpendicular  to  a 
given  segment  at  one  extremity  of  the  segment. 

Take  0  any  point  not  in  segment  AB  and  draw 
a  circle  with  0  as  center  and  OB  as  radius,  cut- 
ting AB  at  D.  Draw  DO  meeting  the  circle  at 
E.    Then  EB  ±  AB  at  B.     Prove  it. 

Ex.  69.  If  chords  AB  and  CD  intersect  at  E  within  the  circle,  prove 
that  A  AEC  and  A  BDE  are  mutually  equiangular. 

Ex.  70.  If  chords  AB  and  CD  extended  meet  outside  the  circle  at 
point  E^  prove  A  ADE  and  A  BCE  are  mutually  equiangular. 

Ex.  71.    Prove  that  the  opposite  angles  of  an  in- 
scribed quadrilateral  are  supplementary. 

(Z  B  is  measured  by  what  ?    Z  2)  ?    .'.  /.  B  +  /.D'}) 


Note.  —  Supplementary  Exercises  9  to  20,  p.  284,  can  be  studied  now. 


118 


PLANE    GEOMETRY  —  BOOK   II 


Proposition  XX.     Theorem 

220.  TJie  angle  formed  hy  a  tangent  and  a  chord 
draivn  to  the  point  of  contact  is  measured  hy  one  half 
its  intercepted  arc. 

A ^..^^ E 


Hypothesis 

Conclusion. 

Proof.     1. 

2. 

3. 

4. 

5. 


Why? 
Why? 

Why? 


AE  is  tangent  to  O  CBB  at  B ;  BC  is  a  chord. 
Z  ABC  is  measured  by  \  BC. 
Draw  diameter  BD\  then  BT)  J_  AE. 
Z  ABD  =  90°,  and  BCD  =  180°. 
.*.  Z  ABD  is  measured  by  \  BCD. 
Z  CBD  is  measured  by  ^  CD. 

Z  ABD  -  Z  CBD  is  measured  by  i  ^OT-  i  CD. 

Ax.  4,  §  51 

6.  .-.  Z  J.5(7  is  measured  by  i  ^O. 

7.  Similarly,  Z.EBC  is  measured  by  i  J3Z>0. 

Ex.  72.  If,  in  the  figure  for  Prop.  XX,  BC  =  llO'^,  how  many- 
degrees  are  there  in  Z  ABC  and  Z  ^50  ? 

Ex.  73.  If  tangents  are  drawn  to  a  circle  at  the  extremities  of  a 
a  chord,  they  make  equal  angles  with  the  chord. 

Ex.  74.  If  two  tangents  drawn  from  a  point  to  a  circle  form  an 
angle  of  60°,  then  each  of  the  tangents  equals  the  chord  joining  the  points 
of  contact.     (Prove  the  triangle  is  equilateral.) 

Ex.  75.  If  a  tangent  be  drawn  to  a  circle  at  the  extremity  of  a  chord, 
the  line  joining  the  mid-point  of  the  intercepted  arc  to  the  point  of  con- 
tact bisects  the  angle  formed  by  the  tangent  and  the  chord. 

Ex.  76.  Prove  that  a  tangent  to  a  circle  at  the  mid-point  of  an  arc 
is  parallel  to  the  chord  of  the  arc.     (§  93.) 


MEASUREMENT   OF  ANGLES  AND  ARCS         119 

Proposition  XXI.     Theorem 

221.  TJie  angle  formed  hy  two  chords  intersecting 
within  a  circle  is  measured  by  one  half  the  sum  of  the 
arcs  intercepted  hy  it  and  its  vertical  angle. 


Hypothesis.     Chords   AB  and    CD   intersect   at  E  within 

OO.  _        _ 

Conclusion.     Z  1  is  measured  by  ^  {AC  +  DB). 

Proof.     1.  Draw  CB. 

2.  Z1  =  Z3  +  Z2.  Why  ? 

3.  Z  3  is  measured  by  J  AG.  Why  ? 

4.  Z  2  is  measured  by  |  DB.  Why  ? 
6.            ,-.  Z  1  is  measured  by  ^  {AC  +  DB)-         Ax.  3,  §  51 

Es.  77.     li  AC=  70°  and  Z)J5  =  50°;  how  many  degrees  are  there  in 
AAECf 

Ex.  78.     liAC=  74°  and  A  AEC  =  50°,  how  large  is  ^  ? 

Suggestion.  —  Let  DB  =  z°. 

Ex.  79.     If  two  chords  intersect  at  right  angles  within  a  circle,  the 
sum  of  one  pair  of  opposite  intercepted  arcs  is  equal  to  a  semicircle. 

B 

Ex.  80.   If  AX  =  (fr  &nd  AB  =  CB,  prove 
A  iif iVJ5  is  an  isosceles  triangle.  „j 

Note.  —  Supplementary  Exercises  21  to  23,  p.  286,  can  be  studied  now 


120  PLANE    GEOMETRY  —  BOOK   II 

Proposition  XXII.     Theorem 

222.  The  angle  formed  hy  tivo  secants  intersecting 
outside  the  circle  is  measured  hy  one  half  the  differ- 
ence between  its  intercepted  arcs. 


Hypothesis.     Secants   AB  and    CD  intersect  at  E  outside 
Q  0.  _         _ 

Conclusion.     Z  ^  is  measured  by  ^  (AC  —  DB). 

Proof.     1.  /.l  =  /.A-\-Z.E.  Why? 

2.  .-.  Z.E  =  Z 1  —  Z  A  By  algebra. 

[Complete  the  proof.     Obtain  the  measures  of  Z  1 
and  Z.A  and  then  determine  the  measure  of  Z  -E".] 

223.  Cor.    1.    The  angle  formed   hy  a     A 
secant  and  a  tangent  is  measured  by  one 
half  the  differeyice  between  its  intercepted 
arcs. 

Prove  Z  ^  is  measured  by 
i  (BC  -  DB). 

224.  Cor.  2.  The  angle  formed  by 
two  tangents  is  measured  by  one  half 
the  difference  between  its  intercepted 
arcs. 

Prove  Z  ^  is  measured  by  ^ 


l(BFD 


BCD). 


MEASUREMENT   OF  ANGLES  AND  ARCS         121 

Ex.  81.     If  in  §  222  J^  =  100°  and  BD  =  40°,  how  large  is  ZE? 

Ex.  82.     K  in  §  222  AC  is  a  quadrant,  and  ZE  is  40°,  how  large  is  BD  ? 
Suggestion.  —  Let  BD  =  z°. 

Ex.  83.  If  AC  in  the  figure  of  Prop.  XXII  is  120°,  and  ZA  =  15°, 
how  large  is  ZE? 

Ex.  84.     If  in   the  figure  for  ^  22S  Z  E  =  50°  and  BD  =  70°,  how 

how  large  is  BFC  ? 

Ex.  85.  If  in  the  figure  for  §  224,  BED  =  |  of  the  circle,  how  Urge 
isZ^? 

Ex.  86.  If  in  §  223  BEG  =  100°,  and  CD  =  200°,  how  many  degrees 
are  there  in  angle  E  ? 

Ex.  87.  If  ^B  is  the  common  chord  of  two  inter- 
secting circles,  and  AC  and  AD  are  diameters  drawn 
from  A,  prove  that  line  CD  passes  through  B. 

Suggestioji.  —  Draw  CB  and  BD,  and  try  to  prove 
CBD  a  straight  line. 

Ex.  88.  A  square  ABCD  is  inscribed  in  a  circle.  A  tangent  is 
drawn  to  the  circle  at  point  A.  How  large  is  the  angle  formed  by  the 
tangent  and  side  AB? 

Ex.  89.  The  line  joining  the  mid-points  of  the  arcs  subtended  by  the 
sides  AB  and  ^C  of  inscribed  A  ABC  cuts  AB  at  F  and  AC  at  G. 
VToyeAF  =  AG. 

Ex.  90.  If  AB  and  AC  are  two  chords  of  a  circle  making  equal 
angles  with  the  tangent  at  A^  prove  AB  =  AC. 

Ex.  91.  If  ABCD  is  an  inscribed  quadrilateral,  and  AD  and  BC  ex- 
tended meet  at  P,  the  tangent  XY  at  P  to  the  circle  circumscribed  about 
the  A  ABP  is  parallel  to  CD. 

Suggestions.  —  1.    XY\\  CD  if  ZDCP  =  ? 

2.    Compare  each  of  these  angles  with  LB  AD.    Recall  Ex.  73. 

Note.  —  Supplementary  Exercises  24  to  30,  p.  286,  can  be  studied  now. 


225.  A  circle  is  said  to  be  inscribed  in  a 
polygon  when  it  is  tangent  to  each  side  of 
the  polygon. 

The  polygon  is  said  to  be  circumscribed 
about  the  circle  ;  as  EFGH, 


122 


PLANE    GEOMETRY  —  BOOK   II 


Pkopositioi^  XXIII.     Problem 
226.   Inscribe  a  circle  in  a  given  triangle. 

A 


Given  A  ABC. 

Required  to  inscribe  a  O  in  A  ABG. 

Construction.     1.   Construct  the  bisectors  BE  and  AD  oi  Z.B 
and  Z  A  respectively,  meeting  at  point  0. 

2.  Construct  OMl^AG. 

3.  With  0  as  center  and  OiJf  as  radius,  draw  a  O. 
Statement.     This  circle  will  be  tangent  to  AB,  BC,  and  AC. 
Proof.     1.    0  is  equidistant  from  the  sides  of  the  triangle. 

§169 

2.  .'.  Js  from  0  to  the  sides  are  all  equal  to  OM. 

3.  .-.  AB,  BG,  and  AG  are  tangents  of  O  0.         §  198 

Note  1.  — Point  0  is  the  point  which  was  called  the  In-center  of  the 
triangle  in  §  169.    The  reason  is  clear  now. 

Note  2.  —  A  circle  can  be  constructed  which  is 
tangent  to  the  sides  AB  and  AC  prolonged  and  to 
BG  as  in  the  adjoining  figure.  It  is  called  an 
Escribed  Circle  and  its  center  is  called  an  Ex- 
center  of  the  triangle. 

There  are  three  ex-centers  for  each  triangle. 

Ex.  92.  Consti-uct  a  triangle  and  construct 
Its  three  escribed  circles. 

Ex.  93.     If  0  is  the  center  of  the  circumscribed  circle  of  A  ABC  and 
OD  is  drawn  perpendicular  to  BC,  prove  Z  BOD  —  /.A. 


MEASUREMENT  OF  ANGLES  AND  ARCS        123 

Proposition  XXIY.     Problem 

227.     I.    Construct  a  tangent  to  a  circle  at  a  point 
on  the  circle. 


Given  O  0  and  point  A  on  it. 

Required  to  construct  a  tangent  to  O  0  at  A. 

Construction  indicated  in  the  figure. 

[Description  and  proof  to  be  given  by  the  pupil.] 

II.    Construct  a  tangent  to  a  circle  from  a  2^oint 
outside  the  circle. 


Given.     O  0  and  point  A  outside  O  0. 

Required  to  construct  a  tangent  to  O  0  from  point  A. 

Construction.     1.   Draw  AO. 

2.  Construct  a  O  on  ^0  as  diameter  intersecting  O  0  at  J5 
and  C. 

3.  Draw  AB  and  AC. 
Statement.     AB  and  AC  are  both  tangent  to  O  O. 

[Proof  to  be  given  by  the  pupil.] 
Suggestion.  —  Draw  OB  and  OC.     Prove  LB  and   Z.  C  are  rt.  A. 


124  PLANE    GEOMETRY  —  BOOK   II 

LOCI 

228.  Illustrative  Problem  1. — Where  are  all  jioints  J 
in.  from  0.? 

Evidently  the  place  of  points  \  in.  from 
0  is  the  circle  with  center  0  and  radius 
iin. 

Instead  of  using  the  word  "  place  '^  it  is 
customary  to  use  the  word  locus  —  a  Latin 
word  meaning  place.  So  the  preceding 
sentence  becomes 

The  locus  of  points  \  in.  from  0  is  the  circle  with  center  0 
and  radius  \  in. 

It  is  evident  that : 

(a)  Every  point  "i  in.  from  0''  is  on  the  circle. 

(6)  Every  point  on  the  circle  is  "  \  in.  from  0.'' 

"  ^  in.  from  0 ''  is  the  condition  which  the  points  satisfy. 

Ex.  94.     Draw  the  locus  of  points  which  are  2  in.  from  a  given  point. 

Ex.  95.  Draw  the  locus  of  the  end  of  a  pump  handle  which  is  33  in. 
long  from  its  end  to  the  point  about  which  it  turns,  if  the  handle  may  be 
moved  through  an  angle  of  100°.     (Let  1  in.  represent  11  in.) 

Ex.  96.     Draw  any  line  of  indefinite  length. 

(a)  Locate  freehand  three  points  above  the  line  which  are  1  in.  from 
the  line. 

(6)  Draw  the  line  which  contains  all  points  which  are  1  in.  from  the 
line  and  lie  above  the  line. 

(c)  Are  there  any  other  points  which  are  1  in.  from  the  line  ? 

(d)  Draw  the  line  showing  where  they  are  to  be  found. 

Ex.  97.  Where  are,  that  is,  what  is  the  locus  of  all  points  on  this 
page  which  are  ^  in.  from  the  left-hand  edge  of  the  page  ? 

Ex.  98.  A  rectangular  lot  is  100  ft.  wide  and  300  ft.  long.  Shrubs 
are  to  be  planted  5  ft.  from  the  lot  line  along  the  two  sides  and  the  back 
of  the  lot.     What  is  the  locus  of  the  shrubs  ? 

Ex.  99.     Draw  two  parallel  lines. 

(a)  Locate  freehand  three  points  which  are  equidistant  from  the  two 
parallels. 

(&)  Draw  the  locus  of  all  points  which  are  equidistant  from  the 
parallels. 


LOCI  125 

Ex.  100.  (a)  Draw  a  line  AB  and  locate  on  it  a  point  C.  Construct 
three  circles,  all  tangent  to  AB  at  C 

(6)  What  is  the  locus  of  the  center  of  a  O  which  will  be  tangent  to  a 
given  line  at  a  given  point  ? 

Ex.  101.     Draw  a  circle  with  radius  1  in. 

Draw  five  radii  of  the  circle.  On  each  radius  locate  a  point  J  in.  from 
the  circle.     Draw  the  locus  of  all  such  points. 

Ex.  102.  Draw  the  locus  of  all  points  outside  a  circle  with  1  in. 
radius  and  \  in.  from  the  circle. 

229.  Def.  If  a  single  geometrical  condition  is  given,  the 
Locus  of  Points  satisfying  that  condition  is  the  line  or  group 
of  lines  such  that : 

(a)  Every  point  in  the  line  (or  lines)  satisfies  the  condition. 
(6)  Every  point  which  satisfies   the  condition  lies   in  the 
line  or  group  of  lines. 

230.  Problem.  Determine  the  locus  of  points  equidistant  from 
tico  given  points.  .^^,„ 

Solution.       (a)     1.    E    is    located    so    that             ^'^ 
EA=:EB.     Similarly,  S  and  Tare  located.  ^ ^ 

2.   Their  position  suggests  that  the  locus  of 
such  points  is  the  _L  bisector  of  AB.  ^"-'^ 

(6)  1.   Assume  that  CD,  the  ±  bisector  ABy  ^:a 

is  the  locus  of  points  equidistant  from  A  and  B. 

2.  Is  every  point  on  CD  equidistant  from  A 

and  B? 

Yes,  by  §  118,  I. 

3.  Is  every  point  equidistant  from  A  and  B  in  line  CD? 

Yes,  by  §  118,  II. 

(c)  .-.  Tlie  locus  of  points  equidistant  from  two  given  points  is 
the  perpendicular-bisector  of  the  segment  between  the  points. 

Note.  —  In  solving  a  locus  problem,  first  locate  three  or  more  points 
satisfying  the  condition  ;  then  decide  what  you  think  the  locus  is  ;  then 
try  to  prove  that  the  supposed  locus  is  the  required  locus.  In  doing  this 
last,  prove  theorems  (a)  and  (b)  of  §  229,  as  is  done  in  part  (6)  of  the 
solution  in  §  230. 


A- 


126 


PLANE   GEOMETRY  —  BOOK  II 


Ex.  103.     Locate  two  points  Xand  F  which  are  2  in.  apart. 

Construct  the  locus  of  points  equidistant  from  X  and  Y. 

Ex.  104.  What  is  the  locus  of  the  vertex  of  an  isosceles  triangle 
which  has  a  given  base  ? 

Ex.  105.  What  is  the  locus  of  the  center  of  a  circle  which  will  pass 
through  two  given  points  ? 

231.   Problem.     Determine  the  locus  of  points  within  an  angle 

which  are  equidistant  from  the  sides  of  the  angle. 

[The  solution  is  to  be  given  by  the  pupil.] 

Suggestions.  —  Model  your  solution  after  that  for  the  problem  in  §  230. 
Recall  §  120,  I  and  II. 

At  the  end  of  your  solution  complete  the  following  sentence  : 
The  locus  of  points  within  an  angle  which  are  equidistant  from  the 
sides  of  the  angle  is  .  .  .  . 

Ex.  106.  Construct  the  locus  of  points  equidistant  from  the  sides  of 
a  right  angle  and  within  the  angle. 

Ex.  107.  What  is  the  locus  of  the  center  of  a  circle  which  is  tangent 
to  the  sides  of  a  given  angle  and  lies  within  the  angle  ? 

Note.  —  For  additional  discussion  of  loci  see  §  238. 


CONSTRUCTION  OF  TRIANGLES 

232.  In  a  A  ABC,  the  sides  op- 
posite angles  A,  B,  and  C  are  marked 
by  the  small  letters  a,  b,  and  c,  re- 
spectively. 

The  letter  h  denotes  an  altitude. 
h^  (read  h  —  sub  —  a)  denotes  the  alti- 
tude to  side  a.     Similarly,  there  are  the  altitudes  h^,  and  h^. 

The  letter  m  denotes  a  median.  The  medians  to  sides  a,  b, 
and  c  are  denoted  by  m^,  m^,,  and  m^  respectively. 

The  letter  t  is  used  to  denote  the  length  of  the  bisector  of  an 
angle  between  the  vertex  and  the  opposite  side.  The  bisectors 
of  angles  A,  B,  and  C  are  denoted  by  t^,  t^,  and  tg. 

Note.  —  This  notation  was  introduced  by  Euler  (1707-1783). 

233.  A  triangle  is  determined  when  three  independent  parts 
are  known. 


CONSTRUCTION  OF  TRIANGLES  127 

General  Suggestions 

1.  Draw  freehand  a  triangle  which  represents  the  desired  Jig- 
iwe,  marking  with  heavy  lines  the  parts  which  correspond  to  the 
given  parts.  Use  this  figure  as  a  guide  in  constructing  the  desired 
triangle.  It  is  rwt  the  desired  triangle,  and  the  parts  marked 
are  not  necessarily  equal  in  size  to  the  given  parts. 

2.  Make  the  construction,  using  the  given  parts. 

3.  Prove  that  the  resulting  triangle  has  all  the  given  parts,  and 
is  the  kind  of  triangle  specified. 

4.  Discuss  the  construction,  determining  whether  there  are  con- 
ditions under  which  it  may  he  impossible  to  construct  a  triangle 
having  the  given  parts.     {See  Prop.  IV,  Book  I,  Discussion.) 

234.  The  following  seven  problems  are  the  fundamental  con- 
struction problems  for  triangles : 

Ejc.  108.     Review  Proposition  IV',  Book  I. 

Ex.  109.  Construct  a  triangle  having  given  two  of  its  sides  and  the 
inchided  angle. 

Suggestion.  —  Let  a  and  6  be  two  given  segments  and  Z  C  a  given  Z,  — 
drawn  at  random  ;  then  construct  the  triangle. 

Ex.  110.  Construct  a  triangle  having  given  two  of  its  angles  and  the 
included  side. 

Discussion.  —  Can  the  triangle  be  constructed  always  : 

(a)  If  both  A  are  acute  ?     (&)  If  both  zi  are  rt.  zl  ?     (c)  If  both  A  are 

obtuse  ?     (d)  If  one  Z  is  obtuse  and  one  is  acute  ? 

Ex.  111.     Construct  a  right  triangle  having  given  its  hypotenuse  and 

a  leg. 

Ex.  112.  Construct  a  triangle  having  given  a  side,  the  opposite  angle, 
and  another  angle. 

Suggestion.  —If  «,  Z^,  and  Z  A  are  given,  then  Z  C  may  be  determined 
by  subtracting  AA+  /.B  from  180°.    Then  A  ABC  can  be  constructed. 

Ex.  113.  Construct  a  right  triangle  having  given  a  leg  and  the  op- 
posite acute  angle. 

Ex.  114.  Construct  a  right  triangle  having  given  the  hypotenuse 
and  an  acute  angle. 

Note.  —  For  further  discussion  of  construction  of  figures  see  §  235 
and  §  241. 


128 


PLANE    GEOMETRY  —  BOOK   II 


Note.  — 
be  omitted 


SUPPLEMENTARY  TOPICS 

The  rest  of  Book  II  is  supplementary  material  and  may  safely 


235.  Triangles  may  be  constructed  when  numerous  other 
combinations  of  three  independent  parts  are  given,  besides 
those  mentioned  already  in  §  234. 

Illustkative  Problem.  —  Construct  a  triangle  having  given 
an  angle,  the  length  of  its  bisector,  and  the  length  of  the  alti- 
tude drawn  from  its  vertex. 

Given 


<^A 


Hence  t\ABE  can  be  con- 


Required  to  construct  A  ABC. 

Analysis.  1.  Let  A^^C,  with  AD  ± 
BC  and  AE  bisecting  A  A  represent  the 
required  figure. 

2.  The  known  parts  are  marked  with 
heavy  lines,  including  Z.BAE  =  AEAG 
=  l/.A. 

3.  A  ADE  is  a  rt.  A  with  a  known  leg 
{—ha)    and   known  hypotenuse    {=1^). 
structed.     (Ex.  111.) 

4.  B  is  on  DE  extended  and  A  BAE  =  J  Z  A 

5.  G is  on  ED  extended  and  Z  EA C  =  ^ZA. 
Construction.     1."  Construct  rt.  A  ADE  with 

leg  =  ha  and  hypotenuse  =  Ia- 

2.  Extend  DE  in  both  directions. 

3.  Bisect  ZA,    and   construct  AB,    making 
Z  EAB  =  ^  Z  ^ ;  let  AB  meet  DE  extended  at  B. 

4.  Construct  AC,  making  Z  EAC  =  ^  ZA,  and  meeting  DE  extended 
at  C. 

5.  Then  A  ABC  is  the  required  triangle. 

Proof.     1.   Z  BAC  =  given  Z  A,  since  it  equals  2(|  Z  A).  Const. 


CONSTRUCTIONS  129 

2.  AD  =  given  ha  and  is  an  altitude  since  Z  D  =  rt.  Z.  Const. 

3.  AE  =  given  tj,  and  is  the  bisector  of  Z  A.  Const. 
Discussion.     The  construction  is  impossible  if  t^  <  ha. 

Note.  —  Observe  that  the  final  triangle  may  appear  quite  different 
from  the  triangle  drawn  for  the  first  step  in  the  analysis. 

236.  Analysis  of  Construction  Problems. 

1.  Draw  a  figure  which  represents  the  desired  figure. 

(a)  Make  this  figure  general.  For  example,  if  a  triangle  is  to  be 
drawn,  do  not  draw  a  right  or  an  isosceles  triangle  unless  such  a  triangle 
is  specified. 

(b)  Remember  that  this  is  not  the  final  figure  and  that  the  parts  in  it 
are  not  necessarily  the  given  parts. 

2.  IVIark  with  heavy  lines  or  with  colored  lines  the  parts 
which  are  known  and  also  those  which  may  be  readily  deter- 
mined from  the  known  parts  by  fundamental  constructions. 

For  example,  if  a  known  line  is  bisected,  then  each  of  the  halves  is 
known. 

3.  Try  to  determine  some  part  of  the  figure  which  can  be 
constructed  by  known  methods.  Usually  this  is  a  triangle. 
This  part  can  usually  be  made  the  basis  for  the  rest  of  the 
construction. 

In  this  connection,  remember  the  first  five  fundamental  triangle  con- 
structions given  in  §  234. 

4.  Try  to  determine  how  the  remaining  parts  can  be  obtained 
from  the  figure  constructed  in  step  3. 

5.  Make  the  construction,  following  the  points  rioted  in  steps 
3  and  4. 

6.  Prove  that  the  resulting  figure  satisfies  the  conditions  of 
the  problem. 

7.  Discuss  the  resulting  figure,  determining,  in  particular, 
whether  the  construction  is  or  is  not  always  possible. 

Note.  —  Systematic  use  of  this  form  of  analysis  is  attributed  to  Plato. 
The  method  of  analysis  has  been  described  as  one  of  the  four  great  steps 
in  mathematics.  Plato  also  introduced  the  restriction  that  constructions 
should  be  made  by  ruler  and  compass  alone. 


130 


PLANE    GEOMETRY  —  BOOK   II 


Illustrative  Problem  1.  —  Construct  A  ABC  having  given 
Z  B,  h^,  and  the  radius  r  of  the  circumscribed  circle. 
Given 


^B= 


Required  to  construct  A  ABC. 

Analysis.  1.  Let  the  adjoining  figure  repre- 
sent the  required  figure. 

2.  A  ABD  is  a  rt.  A,  with  known  leg  ( =  ha) 
and  known  acute  angle  {^B).  .-.  it  can  be  con- 
structed by  Ex,  113. 

3.  Point  0  is  equidistant  from  A  and  B^  a  dis- 
tance equal  to  r.     Hence  O  can  be  located. 

4.  The  circle  can  then  be  drawn,  and  BD^ 
extended,  will  meet  the  circle  at  C. 

Construction  left  to  the  pupil.     Follow  up  the  steps  2,  3,  and  4. 

1.  Construct  a  rt.  ^  with  side  =  ha  and  opposite  Z  =  Z  J5. 

2.  Locate  the  point  O  and  draw  the  circle. 

3.  Extend  BD  and  thus  determine  point  C. 
Proof  and  Discussion  left  to  the  pupil. 


Illustrative  Problem  2. 
tangents  of  two  circles. 

Analysis.     1.    Let  the   circles 
unequal.     Let  the    adjoining 


Construct  the  common  external 


be   assumed 
figure  represent 
the  desired  figure. 

2.  Evidently  ABCO'  is  a  O. 

3.  .•.AB=CO'. 

4.  ..  A0=  OB-  CO'. 

5.  Also,  AO'  and  A' 0'  are  tangents  to  circle  AA'. 

Construction.     1.   Construct  a  circle  with  radius  equal  to  the  difference 
between  the  radii  of  the  two  circles,  and  concentric  with  the  larger  circle. 

2.  Draw  tangents  to  this  circle  from  the  center  of  the  smaller  circle, 
meeting  the  constructed  circle  at  points  A  and  A'. 

3.  Draw  OA  and  OA'  meeting  the  large  circle  at  B  and  B'. 
Complete  the  construction. 

Give  the  Proof  and  the  Discussion- 


CONSTRUCTION  OF  TRIANGLES 


131 


Ex.  115.     Construct  the  common  inter- 
nal tangents  of  two  unequal  circles. 

Constmct  the  triangle  ^5C  having  given  : 

Ex.  116.     b,  c,  he. 

Ex.  117.     a,  c,  and  wig. 

Ex.  118.     a,  6,  and  hg. 

Ex.  119.     6,  he,  B. 

Ex.  120.  Construct  an  isosceles  triangle  having  given  one  base  angle 
and  the  altitude  to  the  base. 

Ex.  121.  Construct  an  isosceles  triangle  having  given  one  side  and 
the  altitude  to  one  of  the  sides. 

Note.  —  Supplementary  Exercises  31-49,  p.  287,  can  be  studied  now. 

Proposition  XXV.     Problem 

237.  Ujoon  a  given  segment  as  chord,  constricct  on 
arc  of  a  circle  such  that  every  angle  inscribed  in  it 
shall  equal  a  given  angle. 


\     N 

Given  segment  AB  and  Z  T. 

Eequired  to  construct  an  arc  upon  AB  as  chord  such  that 
every  angle  inscribed  in  the  arc  shall  equal  Z  T. 

Construction.     1.    Construct  Z  BAC=  Z  T. 

2.  Construct  DE  J_  AB  at  its  mid-point. 

3.  Construct  AFl.  AC  at  A,  intersecting  DE  at  0. 

4.  Construct  a  circle  with  center  0  and  OA  as  radius. 

Statement.     AMB  is  the  required  arc. 

Proof.      (The  pupil  should  now  prove  that  any  Z  AGE  =  Z  T.) 


132  PLANE    GEOMETRY  —  BOOK   II 

FURTHER  DISCUSSION  OF  LOCI* 
238.   Method  of  Attacking  a  Locus  Problem. 

1.  Locate  either  freehand  or  by  construction  three  or  more 
points  which  satisfy  the  given  condition.  These  points 
should  suggest  the  probable  locus. 

2.  Draw  the  probable  locus  and  try  to  prove  that  it  is  the 
real  locus.  To  do  this,  try  to  prove  either  (a)  and  (6)  below, 
or  else  (a)  and  (c) : 

(a)  Every  point  on  the  locus  satisfies  the  given  condition. 
(6)  Every  point  which  satisfies  the  given  condition  lies  on 
the  locus. 

(c)  Every  point  not  on  the  locus  does  not  satisfy  the  given 
condition. 

Note.  —  (6)  is  the  converse  of  (a)  and  (c)  is  the  opposite  of  (a).  The 
opposite  of  (&)  is  : 

(d)  every  point  which  does  not  satisfy  the  given  condition  does  not  lie 
on  the  locus. 

When  (a)  and  (6)  are  known,  then  (c)  and  (d)  can  be  proved  by  the 
indirect  method ;  when  (a)  and  (c)  are  known,  then  (6)  and  (d)  can  be 
proved  in  the  same  manner. 

Illustrative  Problem. — Determine  the  locus  of  the  ver- 
tex of  the  right  angle  of  a  right  triangle  having  a  given  seg- 
ment as  hypotenuse.  ^    q 

Solution.     1.   Let  A  ABC  be  right  tri-        9y^/^^\^<A 
angles  having  the  hypotenuse  AB  and  rt.       ,^^^^^^^^^\V 

z  at  a  ^^— -^=^-« 

2.  This  figure  suggests  that  the  points  C 
lie  on  a  circle  having  AB  as  diameter. 

3.  Assume  that  the  locus  is  the  circle  hav- 
ing AB  as  diameter. 

(a)  Is   every   AAXB,   where    X   is    any 
point  on  the  circle,  a  rt.  A? 

Yes,  since  Z  AXB  is  a  rt.  Z,  by  §  218. 

*  Review  at  this  time  §  229  and  the  two  loci  discussed  in  §  230  and  §  231. 


LOCI  133 

(b)  Is  every  A  A  YB,  where  Y  is  any  point  not  on  the  cir- 
cle, an  oblique  A  ? 

Yes,  since  every  Z  A  YB  is  either  acute  or  obtuse,  according  as  Y  lies 
outside  or  inside  of  the  circle.     (Easily  proved.) 

4.  Hence  the  locus  of  the  vertex  of  the  right  angle  of  a  right 
triangle  having  a  given  segment  as  hypotenuse  is  a  circle  draivn 
on  the  hypotenuse  as  diameter. 

239.   Summary  of  Fundamental  Loci. 

1.  The  locus  of  points  at  a  given  distance  d  from  a  given 
point  0  is  the  circle  drawn  with  0  as  center  and  d  as  radius. 
(§  228.) 

2.  The  locus  of  points  at  a  given  distance  d  from  a  fixed 
line  I  (of  indefinite  length)  is  the  pair  of  parallels  to  I  at  the 
distance  d  from  it.     (Ex.  96.) 

3.  The  locus  of  points  equidistant  from  two  parallel  lines 
is  the  line  parallel  to  them  and  midway  between  them.    (Ex.  99.) 

4.  The  locus  of  points  equidistant  from  two  given  points  is 
the  perpendicular  bisector  of  the  segment  joining  the  points. 
(§  230.) 

5.  The  locus  of  points  equidistant  from  the  sides  of  an  angle 
and  within  the  angle  is  the  bisector  of  the  angle.     (§  231.) 

Cor.     The  locus  of  points  equidistant  from  two  intersecting 
straight  lines  is  the  set  of  bisectors  of  their  included  angles. 
(These  bisectors  form  two  straight  lines.) 

6.  The  locus  of  the  vertex  of  the  right  angle  of  a  right 
triangle  which  has  a  given  hypotenuse  is  a  circle  drawn  upon 
the  hypotenuse  as  diameter.     (§  238.) 

7.  If  A  and  B  are  any  two  fixed  points  and  X  is  a  point 
such  that  Z  AXB  is  a  given  angle,  the  locus  of  X  is  the  arc 
of  a  circle  constructed  upon  AB  as  chord  such  that  every  angle 
inscribed  in  it  equals  the  angle  given.     (§  237.) 

Caution.  —  1.  Remember  that  "  what  is  the  locus  of  ?  "  means  "  what 
is  the  place  of?"  2.  Be  certain  that  you  know  what  "  the  given  con- 
dition" is  in  each  locus  theorem. 


134 


PLANE    GEOMETRY  —  BOOK   II 


240.  Intersection  of  Loci.  —  Sometimes  it  is  specified  that 
a  point  shall  satisfy  each  of  two  given  conditions.  Each  con- 
dition determines  a  locus  for  the  point.  The  point  then  must 
lie  at  the  intersection  of  the  two  loci. 

Illustrative  Problem.  —  Find  all  points  which  are  equi- 
distant from  two  intersecting  lines  and  also  equidistant  from 
two  fixed  points. 

Given  intersecting  lines  AB  and  CD  and  points  B  and  S. 

Required  to  find  all  points  which  are  equidistant  from  AB  and  CD  and 
also  equidistant  from  B  and  S. 

Solution.  1.  The  locus  of 
points  equidistant  from  AB  and 
CD  is  the  set  of  bisectors  of  the 
angles  included  by  them.  (Lines 
^i^and  GH.) 

2.  The  locus  of  points  equi- 
distant from  B  and  S  is  the 
perpendicular  bisector  of  BS. 
(Line  TW.) 

3.  The  required  points  will  be 
at  the  intersection  of  TW  with 
EF  and  GH. 

Discussion.  1.  Usually  there 
are  two  points  ;  as  Xi  and  X2. 

2.  There  may  be  only  one  point,  however,  for  TW  ma,y  be, parallel  to 
one  of  the  lines,  EF  and  GH. 

3.  There  must  always  be  at  least  one  point,  for  T  W  cannot  be  parallel 
to  both  EF  and  GH. 

4.  There  may  be  a  whole  line  full  of  points,  for  TW  may  coincide  with 
EF  or  GH. 

Ex.  122.  In  a  given  line,  find  all  points  which  are  equidistant  from 
two  given  points. 

Ex.  123.  In  a  given  line,  find  all  points  which  are  equidistant  from 
two  given  intersecting  lines. 

Ex.  124.  In  a  given  circle,  find  all  points  which  are  equidistant  from 
two  given  parallel  lines. 

Ex.  125.  Find  all  points  which  are  equidistant  from  two  given  points 
and  also  at  a  given  distance  from  a  given  point. 


LOCI  135 

Ex.  126.  Find  all  points  which  are  equidistant  from  two  given  points 
and  also  at  a  given  distance  from  a  given  line. 

Ex.  127.  Find  all  points  which  are  equidistant  from  two  given  points 
and  also  equidistant  from  two  given  parallels. 

Ex.  128.  Find  all  points  which  are  equidistant  from  two  given 
parallels  and  also  at  a  given  distance  from  a  given  point. 

Ex.  129.  Find  all  points  which  are  at  a  given  distance  from  a  given 
line  and  also  at  another  given  distance  from  a  given  point. 

Ex.  130.  Find  all  points  which  are  equidistant  from  two  parallels 
and  also  equidistant  from  two  intersecting  lines. 

Ex.  131.  Find  all  points  which  are  equidistant  from  two  intersecting 
lines  and  at  a  given  distance  from  a  given  point. 

Ex.  132.  What  is  the  locus  of  the  vertex  of  a  triangle  whose  base 
lies  in  a  given  straight  line  if  the  altitude  to  the  base  is  a  given  segment  ? 

Ex.  133.  What  is  the  locus  of  the  center  of  a  circle  which  shall  be 
tangent  to  a  given  line  and  have  a  given  radius  ? 

Ex.  134.  What  is  the  locus  of  points  at  a  given  distance  from  a 
given  circle  ? 

(The  distance  is  measured  along  a  line  between  the  point  and  the 
center  of  the  circle.) 

Ex.  135.  What  is  the  locus  of  the  center  of  a  circle  which  has  a 
given  radius  and  passes  through  a  given  fixed  point  ? 

Ex.  136.  What  is  the  locus  of  the  center  of  a  circle  which  shall  be 
tangent  to  each  of  two  parallel  lines  ? 

Ex.  137.  What  is  the  locus  of  the  mid-points  of  all  chords  of  a  circle 
that  have  a  given  length  ? 

Ex.  138.  What  is  the  locus  of  the  points  such  that  the  tangents  from 
the  points  to  a  given  circle  shall  have  a  given  length  ? 

Ex.  139.  What  is  the  locus  of  the  mid-points  of  all  parallel  chords 
of  a  circle  ? 

Ex.  140.  What  is  the  locus  of  the  mid-points  of  all  segments  drawn 
from  one  vertex  of  a  triangle  and  terminated  by  the  opposite  side  ? 

Ex.  141.  A  line  AB  of  fixed  length  moves  so  that  A  is  constantly  on 
one  side  of  a  given  right  angle  and  B  is  on  the  other  side  of  the  angle. 
What  is  the  locus  of  the  mid-point  of  the  segment  AB  ?  (Recall  Ex.  175, 
Book  I.) 


136 


PLANE    GEOMETRY  —  BOOK   II 


241.  Construction  of  Figures  by  Intersection  of  Loci. 

Illustrative  Problem.  —  Construct  A  ABC  having  given 
c,  h^,  and  Z  C. 

Given 


he 


from  c.     (Lo- 


Required  to  construct  A  ABC. 

Analysis.  1.  Let  A  ABC  represent  the  de- 
sired triangle,  the  known  parts  being  marked 
by  heavy  lines. 

2.  Line  c  can  be  drawn,  thus  locating  defi- 
nitely points  A  and  B. 

3.  Point  C  is  at  the  distance  h^  from  c.     It 
therefore  lies  on  one  of  two  parallels  to  c  at  the  distance 
cus2,  §239.) 

4.  Point  C  is  such  that  ZACB  must  equal  Z  C.     It  therefore  lies 
on  the  arc  constructed  on  ^B  as  chord, 
the  inscribed  angles  of  which  equal  Z  C. 
(Locus  7,  §239.) 

Construction  is  made  so  as  to  obtain 
the  loci  mentioned  in  steps  3  and  4.  The 
circle  cuts  the  line  BS  at  two  points  (7i 
and  02.  A  ACiB  and  A  AC2B  each  sat- 
isfy the  given  conditions. 

Proof  and  Discussion  left  to  the  pupil. 

Ex.  142.     Given  the  base  and  altitude  of  an  isosceles  triangle,  construct 
the  triangle. 

Ex.  143.     Construct  an  isosceles  triangle  having  given  the  base  and 
the  radius  of  the  circumscribed  circle. 

Ex.  144.     Construct  a  rhombus  having  given  its  base  and  altitude. 

Ex.  145.  Construct  a  right  triangle  having 
given  the  hypotenuse  and  the  length  of  the  altitude 
upon  it. 

Ex.  146.  Construct  an  isosceles  triangle  having 
given  the  base  and  the  angle  opposite  the  base. 


LOCI 


137 


Ex.  147.     Construct  a  triangle  having  given  the  base,  the  altitude,  and 
the  radius  of  the  circumscribed  circle. 

Ex.  148.     Construct  a  triangle  having  given  a  side,  an  adjacent  angle, 
and  the  radius  of  the  circumscribed  circle. 

Ex.  149.     Through  a  given  point  construct  a  circle  with  a  given  radius 
which  shall  be  tangent  to  a  given  line. 

Analysis.     1.   Let  the  circle  with  the  center  C  pass 
through  P  and  be  tangent  to  line  ?. 

2.  O  is  r  distant  from  P.     Hence  it  must  lie  on  a 
circle  having  P  as  center  and  r  as  radius. 

3.  C  is  r  distant  from  I.    Hence  C  must  lie  on  one 
of  two  parallels  to  I  at  the  distance  r  from  I. 

4.  C  must  be  at  the  intersection  of  these  two  loci. 
Construction,  Proof,  and  Discussion  left  to  the  pupil. 

Ex.  150.     Construct  a  circle  with  a  given  radius  which  shall  be  tan- 
gent to  each  of  two  intersecting  lines. 

Ex.  151.     Construct  a  circle  which  shall  be  tangent  to  each  of  two 
intersecting  lines,  tangent  to  one  of  them  at  a  given  point. 

Ex.  152.     Construct  a  circle  through  a  given  point  not  in  a  given  line 
which  shall  be  tangent  to  the  given  line  at  a  given 
point  in  the  line. 


Ex.  153.  Construct  a  circle  having  a  given 
radius  which  shall  be  tangent  to  each  of  two  given 
circles. 


Ex.  154.  A  circular  cylinder  head  12  in.  in  diameter  is  to  have  holes 
bored  in  it  for  12  1-in.  bolts,  equally  spaced  around  the  edge,  with  their 
centers  1^  in.  from  the  edge.  Make  a  scale  drawing  of  the  cylinder  head 
(\"  =  1")  and  mark  the  centers  for  the  12  bolt  holes. 

Ex.  155.  Determine  how  to  construct  the  unit  which  is  repeated  in 
the  design  below.     Construct  it  in  a  circle  of  3-in.  radius. 


Note.  —  Supplementary  Exercises  50  to  63,  p.  288,  can  be  studied  now, 


138 


PLANE    GEOMETRY  —  BOOK   II 


Ex.  156.  Determine  how  to 
construct  the  left-hand  figure. 
Notice  that  it  is  the  basis  for  the 
artistic  design  at  the  right. 
Construct  the  left-hand  figure 
in  a  circle  of  4- in.  diameter. 

Ex.  157.    The   adjoining  figure   indicates   a  form  of  mansard  roof. 
The  chords  AB,   CD,   CE,  and  FG  are  all  ^ 

equal.  ^^^^f^<:^^^ 

Construct  such  a  roof  outline  for  a  build-  ^"C^'  ^^\e 

mg  in  which  AQ  is  30  ft.  and  the  distance 
IM  is  10  ft.     (Let  1  in.  =  5  ft.) 

1&AI  =  IC=  CJ  =  JG'^ 

Is  thelineZ/l|yl(y? 

Ex.  158.     Construct  between  two  parallel  lines  a  set  of  circular  rings 
like  those  in  the  design  below. 


7rV^/^rV/ 


MMM%B 


Design  for  Ornamental 
Stonework  on  a  Bridge 

Ex.  159.  The  adjoining  design  is  a  panel  for 
ornamental  ironwork  on  a  bridge. 

Determine  how  to  construct  the  fundamental  units 
of  the  design,  units  (a)  and  (6). 


Unit  (a) 


Unit  (&) 


Ex.  160.     Construct  a  tangent  to  a  circle  which  will 
be  parallel  to  a  given  line. 

Suggestion.  —  Make  an  analysis  based  on  the  adjoining 
figure. 


MISCELLANEOUS  EXERCISES  139 

Miscellaneous  Exercises 

Ex.  161.  AOB  is  a  diameter  of  O  0.  C  is  any  point  of  AB.  D  is 
the  mid-point  oi  BC  and  E  is  the  mid-point  of  AC.  Prove  Z  DOE  is  a 
right  angle. 

Suggestion. — Draw  CO. 

Ex.  162.  Points  A  and  B  are  on  the  diameter  XY  of  circle  0  at 
equal  distances  from  0.  CA  and  DB  are  perpendicular  to  XF,  meeting 
the  semicircle  at  C  and  D  respectively.  Prove  ABDC 
is  a  rectangle.  . 

Ex.  163.     If  a  circle  is  inscribed  in  a  right  tri-  cl/' 

angle,  the  sum  of  its  diameter  and  the  hypotenuse  is  /J    Or- 

equal  to  the  sum  of  the  legs  of  the  triangle.  /    V     i 

E 

Ex.  164.  If  AB  is  a  common  external  tangent  of  twc^  circles  which 
touch  each  other  externally  at  (7,  prove  Z  ACB  is  a  right  angle. 

Suggestion.  —  Draw  the  common  tangent  of  the  ©  at  C,  meeting  AB  at  D. 

Ex.  165.  Prove  that  the  bisector  of  the  angle  betw^een  two  tangents 
to  a  circle  passes  through  the  center  of  the  circle. 

Suggestions.  —  Draw  radii  to  the  points  of  contact.    Recall  §  120. 

.,„..„.  ».,....„..™. .,...,.. ,^1\ 

sides  of  an  isosceles  triangle  as  diameter  bisects  the     I         /    i    Nl 


Ex.  167.  Two  circles  are  tangent  externally  at  C.  In  one  circle 
A  ABC  is  inscribed,  having  one  vertex  at  the  point  of  contact  6f  the 
circles.  ^Cand  BC  are  extended  through  C,  meeting  the  other  circle 
at  D  and  E  respectively.     Prove  DE  II  AB. 

Suggestion.  —  I)ra.w  the  common  tangent  through  point  C. 

Ex.  168.  If  a  straight  line  be  drawn  through  the  point  of  contact  of 
two  circles  which  are  tangent  externally,  terminating  in  their  circumfer- 
ences, the  tangents  at  its  extremities  are  parallel. 

Suggestion.  —  Draw  the  common  internal  tangent  of  the  circles. 

Ex.  169.  If  AB  and  AC  are  the  tangents  from  point  A  to  the  circle 
O,  Z  BAC  =  2Z0BC. 

Suggestions.  —  1.    Draw  OA.    "What  relation  does  it  bear  to  BC? 
2.    Compare  ZBAO  with  Z  OB C. 


140  PLANE    GEOMETRY  —  BOOK   II 

Ex.  170.  Euclid's  construction  for  the  tangent  to  a  circle  with, 
center  M  from  a  point  A  outside  of  it  is  as  follows : 

1.  Draw  the  circle  with  center  M  and  radius  MA. 

2.  Draw  MA  intersecting  the  given  O  at  B. 

3.  Draw  BC  ±  MA  at  B,  meeting  the  larger  O  at  G. 

4.  Draw  MC^  intersecting  the  given  O  at  D. 

Statement.     AD  is  tangent  to  the  given  O. 
Make  the  construction  and  give  the  proof. 

Ex.  171.  Given  a  side  and  the  diagonals  of  a  parallelogram,  con- 
struct the  parallelogram. 

Ex.  172.  Through  a  given  point  within  a  circle,  construct  a  chord 
equal  to  a  given  chord. . 

Is  there  any  restriction  on  the  location  of  the  point  ? 

A 
Ex.  173.     Construct  a  parallel  to  the  side  BC  of 
A  ABC  meeting  AB  and  AC  &t  D  and  E  respec- 
tively, so  that  DE  will  equal  EC. 

B' 

Ex.  174.  If  point  B  bisects  arc  AC  of  a  circle,  then  ZA  of  A  ABC 
equals  Z  C. 

Ex.  175.  Prove  that  the  bisectors  of  the  angles  of  a  circumscribed 
quadrilateral  pass  through  a  common  point. 

Ex.  176.  Prove  that  two  chords  which  are  perpendicular  to  a  third 
chord  at  its  extremity  are  equal. 

Ex.  177.     If    XA  =  YG   and   BG  =  AB,    prove 
A  AXB  ^  A  YCS. 

Ex.  178.     In  the  figure  for  Ex.  177  draw  A  C  cutting 
XB  at  M  and  YB  at  N.      Prove  A  AXM^  A  YCJ^. 

Ex.  179.  A  carpenter  has  a  tool  called  a  gauge  which  illustrates  and 
applies  one  of  the  fundamental  loci  theorems. 

The  shaded  rectangle  represents  the  end  of  a  board  ;  the  tool  is  upon 
the  right-hand  side  of  the  board.  P  is  a  marking  point  which  extends  to 
the  under  side  of  the  tool.  AB  is  &  movable  part  which  can  be  fixed  at 
any  short  distance  from  P  by  means  of   a  p  ^ 

screw  at  A.     By  moving  the  gauge  so  that    '■^^^^^^gLgg;:^:^;-::^^^  J 
AB  is  constantly  against  the  edge  of  the 
board,   the  point  P  traces  upon  the  upper  side  of  the  board  a  line 
parallel  to  the  edge  of  the  board.     Why  is  this  so  ? 


lY 


B 


BOOK  III 
PROPORTION  —  SIMILAR  POLYGONS 

242.  The  Eatio  of  one  number  to  another  is  the  quotient  of 
the  first  divided  by  the  second. 

Thus,  the  ratio  of  a  to  &  is  -  ;  it  is  also  written  a:  b. 
h 

The  numerator  is  called  the  Antecedent  and  the  denomina- 
tor is  called  the  Consequent. 

Since  a  ratio  is  a  fraction,  it  is  subject  to  the  usual  rules 
for  operations  with  fractions. 

243.  The  ratio  of  two  concrete  quantities  of  the  same  kind  is 
the  ratio  of  their  measures  in  terms  of  a  common  unit.  (§  212.) 

Thus,  the  ratio  of  350  lb.  to  2  tons  is  ^^  or  ^. 

Ex.  1.     Express  the  following  ratios  in  their  simplest  form. 

(a)  3  to  9.        (c)  6x  to  2  x.  (e)  f  to  y\.      (g)  25  to  375. 

(6)  12  to  2.      (d)  6  a2  to  15  a^.     (/)  ^  to  ].      {h)  a^  -  b'^  to  a^  -  b\ 

Ex.  2.  A  line  15  in.  long  is  divided  into  two  parts  which  have  the 
ratio  2  :  3.     Find  the  parts. 

Suggestion.  —  If  the  short  part  contains  x  in.  and  the  long  part  (15  —  a;)  in., 

X  2 

then  =  -•    Complete  the  solution. 

15  — a;     3 

Ex.  3.     Divide  a  line  63  in.  long  into  two  parts  whose  ratio  is  3  :  4. 

Ex.  4.  Divide  36  into  two  parts  such  that  the  ratio  of  the  greater 
diminished  by  4  to  the  less  increased  by  3  will  be  3  : 2. 

Ex.  5.  The  ratio  of  the  height  of  a  tree  to  the  length  of  its  shadow 
on  the  ground  is  17  :  20.  Find  the  height  of  the  tree  if  the  length  of 
the  shadow  is  110  feet. 

Ex.  6.  What  is  the  ratio  of:  (a)  a  right  angle  to  a  straight  angle? 
(6)  a  right  angle  to  the  perigon  ?  (c)  one  angle  of  an  equilateral  tri- 
angle ^to  the  sum  of  all  the  angles  of  the  triangle  ?  (d)  one  side  of  a 
square  to  the  perimeter  of  the  square  ? 

141 


142  PLANE    GEOMETRY  —  BOOK   III 

244.  A  Proportion  is  a  statement  that  two  ratios  are  equal  j 

as,  -  =  -,  ov  a:o  =  c:  a. 
b      d 

This  proportion  is  read  "  a  is  to  6  as  c  is  to  d." 

1      5 
Thus,  1,  3,  5,  and  15  form  a  proportion  since  -  =  ^. 

S       15 

This  means  that  1  bears  to  3  the  same  relation  that  5  bears  to  15. 

The  first  and  fourth  terms  of  a  proportion  are  called  the 
Extremes,  and  the  second  and  third  terms,  the  Means. 

In  the  proportion  a  :  b  =  c  :  d,  a  and  d  are  the  extremes  and  b  and  c 
are  the  means  ;  a  and  c  are  the  antecedents,  and  b  and  d  are  the  con- 
sequents. 

Ex.  7.  Select  four  numbers  which  form  a  proportion  like  the  arith- 
metical illustration  in  §  244. 

Ex.8.     (a)Is^  =  ^?  (b)ls^  =  ^?  (c)Is?zz:A? 

Ex.  9.  Find  the  value  of  the  literal  number  in  each  of  the  following 
proportions. 

^^48  ^M6      8  ^^2  3 

(^,)  L0  =  2.  1^3.  3+^  ^5. 

Ex.  10.     Find  the  value  of  x  in  each  of  the  following  proportions. 

^   ^   b      X  ^  ^   36      2c  ^       s        t  P        X 

245.  Proportion  is  used  in  a  great  variety  of  ways. 

Example.  —  The  cost  of  a  number  of  articles  of  a  given 
kind  is  ''proportional"  to  the  number  of  articles. 

Thus,  the  cost  of  seven  books  is  to  the  cost  of  three  books  of  the  same 

kind  as  7  is  to  3.     Hence,  if  3  books  cost  .§1.35,  the  cost  of  7  books  maj'- 

X        7 

be  determined  from  the  proportion  =  -  • 

^    ^  1.35      3 

Ex.  11.     Determine  by  proportion  the  cost  of  13  yd,  of  cloth  if  the 

cost  of  5  yd.  of  the  same  cloth  is  70^. 

Ex.  12.  Determine  by  proportion  the  distance  an  automobile,  will 
travel  in  one  hour  if  it  travels  2  mi.  in  5  minutes. 


PROPORTION  143 

Ex.  13.  If  a  girl  makes  3  14.25  profit  from  15  hens  in  one  year,  what 
profit  can  she  expect  from  60  hens,  assuming  the  same  average  profit  per 
lien  ? 

246.  The  Fourth  Proportional  to  three  numbers  a,  6,  and  c 
is  the  number  x  in  the  proportion  a:  b  =  c:  x. 

2  4 
Thus,  the  fourth  proportional  to  2,  3,  and  4  is  the  number  x  in  -  =  - . 

3  X 
.-.  2x  =  12,  orx  =  6. 

Note.  —  The  numbers  must  be  placed  in  the  proportion  in  the  order  in 
which  they  are  given  as  in  the  illustrative  example. 

Ex.  14.     Find  the  fourth  proportional  to  : 

(a)  2,  3,  and  4.  (d)  25,  15,  and  10. 

(6)  3,  2,  and  4.  (e)   3  a,  2  6,  and  c. 

(c)  4,  3,  and  2.  (/)  r,  rs,  and  s. 

247.  The  Third  Proportional  to  two  numbers  a  and  b  is  the 

number  x  in  the  proportion  a:  b  =  b  :  x. 

2      3 
Thus,  the  third  proportional  to  2  and  3  is  a;  in  -  =  -  • 

•  9      ^      ^ 

.-.  2x  =  9,  orx  =  -. 

2 

Ex.  15.     Find  the  third  proportional  to  : 

(a)  3  and  5.  (&)  2  and  5  t.  (c)  5  and  10. 

248.  A  Mean  Proportional  between  two  numbers  a  and  b  is 
the  number  x  in  the  proportion  a:  x  =  x:  b. 

Example.  —  A  mean  proportional  between  2  and  3  is  a;  in : 

2^x 
X     Z' 
.'.  a;2  =  6,  orx=±V6. 

There  are  two  mean  proportionals  between  any  two  num- 
bers. The  positive  one  is  implied  when  "the"  mean  propor- 
tional is  specified. 

Ex.  16.    Find  the  mean  proportional  between : 

(a)  75  and  12.  (c)  2  r^t  and  IS  rt. 

(6)  3  a  and  a.  (d)  6f  and  |. 


144  PLANE    GEOMETRY  —  BOOK   III 

FUNDAMENTAL  THEOREMS   OF  PROPORTION 

249.  The  mean  proportional  between  two  numbers  is  the  square 
root  of  their  product. 

For  the  mean  proportional  between  a  and  6  is  a;  in  : 
a  _x 
X      b 
.'.  x'^  =  ab;  or  ic  =  Vab. 

250.  In  a  proportion,  the  product  of  the  extremes  is  equal 
to  the  product  of  the  means. 

If  -  =  -,  then  ad  =  be. 

b      d 

Suggestion,  —  Multiply  both  members  of  the  proportion  by  bd. 

2      6 
Example.— Since  -  =  -,2x9  should  equal  3x6.     Does  it?- 

251.  If  three  terms  of  one  proportion  are  equal  respectively  to 
the  three  corresponding  terms  of  another  proportion,  the  fourth 
terms  also  are  equal. 

If  -  =  -  and  -  =  -,thena:  =  2/. 

b     x  by 

Suggestion.  —  Determine  x  from  the  first  proportion  and  y  from  the  second. 

Example.  —  If  -  =  -  and  -  =  -,  then  x  should  equal  y. 
3      X  3     y 

252.  If  the  product  of  two  numbers  is  equal  to  the  pi^oduct  of 
two  other  numbers,  one  pair  may  be  made  the  means  of  a  propor- 
tion having  the  other  pair  as  the  extremes. 

If  mn  =  xy,  then  —  =  ^ , 

X      n 

Proof.     Dividing  both  members  of  the  given  equation  by  xn, 

mn  _xy         m  _y_ 
xn      xn         X      n 

3  4 

Example.  —  Since  3x8  =  6x4,  -  should  equal  - .   Does  it? 

6  8 


PROPORTION  145 

Ex.  17.     If  mn  =  xy,  prove  : 

(a)   ^=^.      (Divide  by  yn.)  (6)  ^  =  ^'.  (c)  ^  =  i^. 

^y?i  my  X      m 

Ex.  18.  Since  4  x  5  =  2  x  10,  write  four  proportions  which  involve 
4,  6,  2,  and  10. 

253.  I7i  any  proportion,  the  terms  are  in  proportion  by  Alter- 
nation ;  that  is,  the  first  term  is  to  the  third  as  the  second  is  to 
the  fourth. 

If  ^  =  ^,then«=^. 

b     d  c      d 

Proof.     1.    Since      -  =  -,  then  ad  =  be.  §  250 

b     d 

2.   Since  ad  =  be,  then  -  =  - .  §  252 

c      d 

Example.  —  Since  -  =  — - ,  then  -  should  equal  — .     Does  it  ? 
6     12'  4  ^12 

Ex.  19.     Write  each  of  the  following  proportions  by  alternation : 

^  ^    S      15'     ^^5      r'     ^       s      y 

254.  Ill  any  proportion,  the  terms  are  in  proportion  by  In- 
version ;  that  is,  the  second  term  is  to  the  first  as  the  fourth  is  to 
the  third. 

If  «  =  ^,then  *  =  5?. 

b     d  a      c 

Proof.     1.    Since      -  =  - ,  then  ad  =  be.  Why  ? 

b      d 

2.   Since  ad  =  be,  then  -  =  -  •  Why  ? 

a     c 

O  A  fi  "I  O 

Example.  —  Since -  =  — ,  then  -  should  equal  — .  Does 
it?  6     12'  2  ^4 

Ex.  20.     Write  each  of  the  three  proportions  of  Ex.  19  by  inversion. 

Ex.  21.     Write  the  proportion  f  =  -,  by  inversion,  and  then   write 

h     y 

the  resulting  proportion  by  alternation. 


146  PLANE   GEOMETRY  —  BOOK  III 

255.  lyi  any  proportion,  the  terms  are  in  proportion  by  Com- 
position ;  that  is,  the  sum  of  the  first  two  terms  is  to  the  second, 
as  the  sum  of  the  last  two  terms  is  to  the  fourth. 

If  «  =  :?,then^i^  =  ^+^. 

b      d  b  d 

Proof.     1.   Since  -  =  -,  then  ^+1=^  +  1.  Why  ? 

b      d  b  d 

2.  .*.  — ^^  =  -^ — .  Algebraic  addition. 

b  d 

Example.     Since  -=  — ,  then  -^ —  should  equal  — ^t — ^^ 
6      12'  6  ^  12 

Does  it  ? 

Ex.  22.     Write  the  three  proportions  of  Ex.  19  by  composition. 

Ex.  23.     Write  the  proportion  -  =  -  : 

b     y 

(a)  By  composition  and  the  result  by  inversion. 

(6)  By  inversion  and  the  result  by  composition. 

(c)  By  composition  and  the  result  by  alternation. 

(d)  By  alternation  and  the  result  by  composition. 

256.  In  any  proportion,  the  terms  are  in  proportion  by  Divi- 
sion; that  is,  the  first  term  minus  the  second  is  to  the  second,  as 
the  third  minus  the  fourth  is  to  the  fourth. 

Tu  a     c,i        a  —  b      c  —  d 

If  -  =  -,  then  = 

b     d  b  d 

Proof.     1.    Since  -  =  -,  then  ^-1  =  ^-1.  AVhy  ? 

b      d  b  d 

(Complete  the  proof.) 
Ex.  24.     Write  the  three  proportions  of  Ex.  19  by  division. 

Ex.  25.     Write  the  proportion  -  =  - : 

b     y 

(a)  By  division  and  the  result  by  alternation. 

(&)  By  alternation  and  the  result  by  division. 

(c)  By  inversion  and  the  result  by  division. 

{d)  By  division  and  the  result  by  inversion. 


PROPORTION  147 

257.  In  any  proportion^  the  terms  are  in  proportion  by  Com- 
position ayid  Division ;  that  is,  the  first  term  plus  the  second  is  to 
the  first  term  minus  the  second,  as  the  third  term  plus  the  fourth 
is  to  the  third  term  minus  the  fourth. 

If  ?  =  «,then^Jl^  =  <l±^. 

h      a  a  —  b     c  —  d 

Proof.     1.   Since  «  =  ^,  then  ^^i^  =  ^^t_^-.  §255 

b     d  b  d 

2.   Also  a-±^c^-_d^  ^  256 

b  d 

3  •    c^4-&  ,  a  —  b  _c-\-  d  .  c  —  d  Why? 

"      b      '      b  d      '      d     '  ^  ' 

J,  a  -\-b         b  c-\-d        d 


or 


a  —  b         d 

a+  b  _c +d 
a— b     c—d 


^  o-        10     15  ^,        10+2     15  +  3     T^       .,., 

Example.  —  Since  -—  =  — ,  then  ^^         =  ^^  '      .    Does  it  .'* 
2       3  '  10  -  2      15  -  3 

Ex.  26.     Write  the  proportion  -  =  - : 

h     y 

(a)  By  composition  and  division  and  the  result  by  inversion. 

(6)  By  inversion  and  tiie  result  by  composition  and  division. 

(c)  By  composition  and  division  and  the  result  by  alternation. 

(d)  By  alternation  and  the  result  by  composition  and  division. 

258.  In  any  jyroportion,  like  powers  or  like  roots  of  the  terms 
are  in  proportion. 

If  ^  =  ^,  then  ^  =  ^,  and  also  ^5  =  ^. 
b     d'  b^     d-'  yi      ^ 

The  first  of  these  conclusions  follows  from  raising  both  members  of  the 
given  proportion  to  the  nth  power,  and  the  second  from  taking  the  nth 
roots  of  both  members. 

Ex.  27.   n«  =  ^, prove:  (a)  m=,n^.  a)  r^  =  ^. 

b     d  b       d  mb     d 


148  PLANE    GEOMETRY  —  BOOK   III 

259.  The  preceding  paragraphs  give  some  of  the  essential 
facts  about  the  subject  of  proportion.  Remember  that  the 
terms  in  every  case  are  numbers.  We  shall  be  dealing  with 
ratios  and  proportions  of  geometrical  magnitudes.  However, 
as  explained  in  §  243,  we  replace  the  magnitudes  themselves 
by  their  measures  in  terms  of  common  units  so  that  the  terms 
are  again  numbers.  The  consequence  is  that  the  theorems 
about  proportions  all  apply  to  the  proportions  which  we  shall 
encounter. 

Thus,  if  AB,  CD,  EF,  and  OH  are  four  segments  such  that 

4^  =  RZ   then  ABx  GH=EFx  CD. 
CD      OH' 

This  means  that  the  product  of  the  numerical  measures  of  AB  and 
dr  equals  the  product  of  the  numerical  measures  of  ^i^'and  CD. 

A  similar  interpretation  must  be  given  to  all  applications  of  §  246  to 
§  258  inclusive. 

PROPORTIONAL  LINE-SEGMENTS 

260.  Introduction.      If    AJEJ  =  EB     ,       a    E 

and  CF=FD,  then  ^=^   since    ^        //       f ^ 

EB     FD  ^ ' • ^ 

each  ratio  equals  1.     Again,  in  the  same  figure,  if  G  bisects 

AC      1  PIT      1 

AE  and  H  bisects  CF,  then  ^=i  and  also  ~^  =  ^',  hence 

GB     3  HD     3 

AO      OTT 

'^=-^  =  — —   This  means  that  AG  bears  to  GB  the  same  relation 

GB     HD 

that  CH  bears  to  HD.  G  and  H  are  said  to  divide  AB  and 
CD  proportionally. 

Def.  Two  line-segments  are  divided  proportionally  when  the 
segments  of  one  have  the  same  ratio  as  the  corresponding  seg- 
ments of  the  other. 

Ex.  28.  Draw  a  A  ABC,  having  AB  =  2  in.,  AC  =  3  in.,  and  BC 
=  4  in.  Place  X  on  AB,  so  that  AX=  .5  in.  Draw  from  X  a  parallel 
io  BC  meeting  ^C  at  Y. 

(a)  Measure  A  Y  and  YC,  and  determine  the  ratio  of  A  Y  to  YC. 

(6)   What  is  the  ratio  of  ^Xto  XB  ? 

(c)  Do  the  sides  appear  to  be  divided  proportionally  ? 


PROPORTION 

Proposition  I.     Theorem 


149 


261.  A  parallel  to  one  side  of  a  triangle,  intersecting 
the  other  two  sides,  divides  the  other  two  sides  pro- 
portionally. 


Hypothesis.  In  A  ABC,  DE  II  BC,  meeting  AB  at  D  and 
AC  at  E. 

Conclusion.  ^  =  ^. 

DB     EC 

Case  I.     Suppose  that  AD  and  DB  are  commensurable. 

§211 
Proof.     1.   Let  AF  be  a  common  measure  contained  4  times 
in  AD  and  3  times  in  DB. 

"  DB     3  * 

2.  Draw  lis  to  BC  through  the  points  of  division  on  AB. 
Then  AC  will  be  divided  into  equal  segments,  of  which  4  are 
in  AE  and  3  are  in  EC.  §  147 

3  .    AE^^ 

"EC     3* 

AD     AE 


4.   Then,  from  steps  (1)  and  (3), 


DB     EC 


Ax.  1,  §  51 


Case  II.  Suppose  that  segments  AD  and  DB  are  incom- 
mensurable. §  211 

The  proof  given  for  Case  I  will  not  apply,  as  no  common 
measure  with  which  to  divide  both  AD  and  DB  can  be  found. 

The  theorem  is  true  hoivever  for  the  incommensurable  case 
also.  The  proof  is  given  in  §  424  and,  if  desired,  may  be  read 
at  this  time. 


150 


PLANE    GEOMETRY  —  BOOK   III 


262.   Cor.  1.    Since  AD  :  DB 


AD  4-  DB     AE  +  EC        AB 

' — ' or 


Why 


AE :  EC,  then 

AC 

DB  EC      '  '    DB     EG 

That  is,  one  side  is  to  its  lower  segment  as  the  other  side 
is  to  its  lower  segment. 

A 


AD 


^,  then  m 
EC  AD 


EG 

ae' 

AB     AC 


or  = 


Why? 
Why? 


263.  Cor.  2.    Since 

DB 

DB -\- AD  ^EC  +  AE 
"AD  AE      '       AD     AE 

That  is,  one  side  is  to  its  upper  segment  as  the  other  side 
is  to  its  upper  segment. 

264.  Numerous  other  proportions  may  be  derived  from  the 
proportions  obtained  in  §§  261-2G3  by  making  allowable 
changes  in  them. 

From  §261,  AD:  AE  =^  DB:  EC.  Why? 

Erom  §  262,  BD:AB  =  EC:  AC  Why  ? 

From  §  263,  AD:AB  =  AE :  AC  Why  ? 

Note.  —  In  every  case,  corresponding  segments  occur  in  the  proportion 
in  the  same  manner. 

265.  For  convenience,  reference  may  be  made  to  any  of  the 
proportions  developed  in  §§  261-264  by  quoting  the  authority: 

A  parallel  to  one  side  of  a  triangle  divides  the  other  two  sides 
proportionally. 

Ex.  29.  If,  in  the  figure  of  §  261,  AD  is  i  of  BD,  what  is  the  ratio  of 
AE  to  EC? 

Ex.  30.  liAD  =  S  in.,  DB  =  6  in.,  and  EC  =  6  in.,  find  AE. 

Ex.  31.  If  AB  =  12  in.,  AC  =  15  in.,  and  AE  =  6  in.,  find  AD. 


PROPORTION  151 

Proposition  II.     Problem 

266.    Construct  the  fourth  proportio'nal  to  three  given 
segments. 


Given  segments  m,  n,  and  p. 

Required  to  construct  the  fourth  proportional  to  m,  w,  and  p. 
Analysis.     1.   Let  x  represent  the  fourth  proportional. 
Then  m  :  n  =p  :  x. 

2.   This  suggests  the  following  construction. 
Construction.     1.   On  side  AB  of  a  convenient  angle,  Z  BAC, 
take  AD  =  m,  and  DE  =  w ;  on  ^C,  take  AF=2). 

2.   Draw  DF  and  construct  EG  \\  DF,  meeting  AC  at  G. 
Statement.     Then  FG  is  the  fourth  proportional  to  m,  n, 

and  p. 

[Proof  to  be  given  by  the  pupil.] 

267.   Cor.    Construct  the  third  proportional  to  m  and  n. 
Analysis.     1.    Let  x  represent  the  third  proportional  to  m 
and  71. 

2.    Then  m:n  =  n:  x. 

[Construction  and  proof  to  be  given  by  the  pupil.] 

Ex.  32.  Construct  the  fourth  proportional  to  segments  which  are  2  in., 
1  in.,  and  3  in.  in  length.  Measure  the  resulting  segment.  Verify  your 
work  by  computing  the  fourth  proportional  to  2,  1,  and  3,  as  in  Ex.  14. 

Ex.  33.  Let  OB  be  any  line  within  ZAOC  and  Fand  V  any  two 
points  on  OB.  Let  FXand  yX'  be  perpendiculars  to  OA,  and  TZ  and 
Y'Z'  be  perpendiculars  to  OC.    Prove  that  OX  :  OX'  =  OZ  :  OZ'. 

Ex.  34.    A  line  drawn  parallel  to  the  bases  of  a  b c 

trapezoid  and  intersecting  the  non-parallel  sides,  A  \ 

divides  the  non-parallel  sides  proportionally.  ^^ — '-^ ^^ 

Prove  BE:EA=CF:  FD. 


A^ k 'D 


152  PLANE    GEOMETRY  —  BOOK   III 

Proposition  III.     Theorem 

268.    A  line  ivhich  divides  tivo  sides  of  a  triangle  pro- 
portionally is  parallel  to  the  third  side. 

A 


Hypothesis.     In  A  ABC,  DE  intersects  AB  and  AC  so  that 

AB^AC 

AD  Ae' 
Conclusion.  DE  \\  BC. 

Proof.    1.   Assume  DF  \\  BC,  meeting  AC  at  F. 

4.  .-.  AF=AE.  §  251 

5.  .*.  i^  coincides  with  E,  and  DE  with  DF.  Why  ? 

6.  .'.  DE  \\  BC  Step  1 

Ex.  35.  If  ^D  =  3  in.,  AB  =  12  in.,  A0=  10  in.,  and  AE  =  2.5  in., 
is  DE  li  BC? 

Ex.  36.  It  AD  =  6  in.,  BD  =  10  in.,  AE  =  6  in.,  and  EC  =  11  in.,  is 
DE  II  BC? 

Ex.  37.     What  relation  is  tliere  between  Prop.  I  and  Prop.  Ill  ? 

269.   Def.     If  P  is  a  point  of  segment 

AB,  then  P  divides  AB  internally  into  two  a £ b 

segments  AP  and  PB. 

Ex.  38.  Construct  a  A  ABC  having  AB  =  2  in.,  BC  =  4  in.,  and 
-4(7  =  4.5  in.  Let  the  bisector  of  Z  J5  meet  ^O  at  D.  Measure  AD  and 
DC.     Compare  the  ratio  of  AB  to  BC  with  the  ratio  of  AD  to  DC. 

Note.  —  Supplementary  Exercises  1  to  3,  p.  289,  can  be  studied  now. 


SIMILAR   POLYGONS 


153 


Proposition  IV.     Theorem 

270.  In  any  triangle,  the  bisector  of  an  interior  angle 
divides  the  opposite  side  internally  into  segments  pro- 
portional to  the  adjaceiit  sides  of  the  triangle. 


Hypothesis. 
Conclusion. 


2.   Then,  in  A  EBC, 


AD  bisects  Z  A  oi  A  ABC,  meeting  BC  at  D. 

BD^BA 

DC     AC 

Proof.     1.    Draw  BE  II  DA,  meeting  CA  extended  at  E. 

BD      EA  ^,     o 

_  =  _.  Why? 

Prove  now  that  BA  =  EA  and  substitute  it  for  EA  in  step  2. 

Suggestions.  —  (1)  Recall  §  123.  (2)  Compare  Z  1  aud  Z  3  with  Z  5  and  Z  4 
respectively,  and  use  the  hypothesis. 

Ex.  39.  The  sides  of  a  given  triangle  are  10,  20,  and  12  inches  re- 
spectively. Find  the  segments  of  the  side  of  length  12  in.  made  by  the 
bisector  of  the  angle  opposite  it. 

Ex.  40.    The  sides  of  a  triangle  are  6,  7,  and  8  inches  respectively. 
Find  the  segments  of  each  side  made  by  the  bisector  of  the  opposite  angle. 
Note.  —  Supplementary  Exercises  4  to  5,  p.  289,  can  be  studied  now. 


SIMILAR  POLYGONS 

271.  Introduction.     The    triangles    below   are    similar   tri- 
angles.    Notice  that  they  appear  to  have  the  same  shape. 


154  PLANE    GEOMETRY  —  BOOK   III 

Ex.  41.  Construct  a  A  ABC  having  AB  =  2  in.,  Z^  =  50°,  and 
ZB  =  80°.  Also  construct  a  A  A'B'  C,  having  A'B'  =  iin.,  ZA'  =  50°, 
and  ZB'  =  80°. 

(a)  Do  the  triangles  appear  to  be  similar,  in  the  sense  that  they  have 
the  same  shape  ? 

(&)  Determine  the  lengths  of  BG,  B'C,  AC,  and  A'C.  Then  deter- 
mine the  approximate  values  of  the  ratios:  ABiA'B';  BC:B'C'i 
AC:  A'C. 

(c)  Do  the  ratios  appear  to  be  about  equal  ? 

272.   Def.     Two  polygons  are  similar  (~)  if: 

(1)  Their  homologous  angles  are  equal ; 

(2)  Their  homologous  sides  are  proportional. 


E'  D' 


Thus,  ABCDE^A'B'C'D'E'  if : 

(1)  ZA=ZA'',  ZB  =  ZB'',  ZC^ZO',  etc.  and 

(2)^=^  =  ^  =  ...etc. 
^  ^  A'B'      B'C      CD' 

The  ratio  of  any  two  homologous  sides  of  two  similar  poly- 
gons is  called  the  Ratio  of  Similitude  of  the  polygons. 

Note.  —  Two  polygons  may  have  their  homologous  angles  equal  and 
still  fail  to  be  similar  ;  as  a  square  and  a  rectangle. 

Ex.  42.     Are  two  squares  similar  ?    Why  ? 

Ex.  43.    Are  two  equilateral  triangles  similar  ?     Why  ? 

Ex.  44.     Are  two  rectangles  necessarily  similar  ? 

Ex.  45.  The  sides  of  one  triangle  are  1  in.,  1.5  in.,  and  2  in.  respec- 
tively. The  shortest  side  of  a  similar  triangle  is  2  in.  What  are  the  other 
sides  of  the  second  triangle  ?     Construct  the  two  triangles. 

Ex.  46.  The  sides  of  one  pentagon  are  3,  4,  5,  8,  and  11  in.  respec- 
tively. The  shortest  side  of  a  similar  pentagon  is  9  in.  How  long  are  the 
remaining  sides  of  the  second  pentagon  ? 


SIMILAR   POLYGONS 


155 


Proposition  V.     Theorem 

273.    Two  triangles  are  similar  if  they  are  mutually 
equiangular. 


Hypothesis.     In  A  ABC  and  AXFZ: 

/:a  =  z.X',  zb^^Y',  zc=z.z. 

Conclusion.  A  ABC  ^  AXYZ. 

Plan.     We  must  prove  the  homologous  sides  proportional. 

Proof.  1.  Place  AXYZm  the  position  ADE,  Z.  X  coincid- 
ing with  Z  Ay  and  vertices  y  and  Z  falling  at  D  and  E  respec- 
tively. 

2.    Since         Z  ADE=  Z  B,  then  DE  |j  BG.  Why  ? 

.-.  4^  =  ^^  Why? 

AD     AE  ^ 

AB^AC 

XY     XZ 

5.  By  placing  A  XZY  so  that  Z  Y  coincides  with  its  equal 
Z  B,  it  may  be  proved  that 

AB      BC 


3. 


4.   That  is 


XY     YZ 

6.   From  steps  (4)  and  (5),  11=  11=  ||. 


7. 


AABC^AXYZ. 


Why? 
,  §272 


274.  Cor.  1.      Tivo  triangles  are  similar  if  two  angles  of  one 
are  equal  respectively  to  two  angles  of  the  other. 

Suggestion.  — Recall  §  111. 

275.  Cor.  2.     Two  right  triangles  are  similar  if  an  acute 
angle  of  one  is  equal  to  an  acute  angle  of  the  other. 


156 


PLANE    GEOMETRY  —  BOOK   III 


276.     Cor.  3.     Two  triangles  are  similar  if  their  sides  are 
parallel  each  to  each. 


Hypothesis.    In  A  ABG  and  (either)  A  A'B'C : 
AB  II  ^'i3';  BG  II  B^O  ;  AC  \\  A'C. 

Conclusion.  A  ABC  ^  A  A'B'C. 

Suggestion.  —  Recall  §  105. 

277.   Cor.  4.     Two   triangles  are  similar  if  their  sides  are 
perpendicular  each  to  each.  ^ 

Hypothesis.     In  A  ABC  and  A  A'B'C ; 

AB±A'B';  BC±B'C'',  AC ±  A'C. 
Conclusion.     A  ABC  ~  A  A'B'C,  ^ 

Ex.  47.     Construct  any  triangle  ABC.    Upon  a  segment  XY  which 
equals  2  AB  construct  a  triangle  similar  to  A  ABC.  (§  274) 

Ex.  48.     If  X  and  Y  are  any  two  points  on  the  side  ^  ^^ 

BC  ot  acute  angle  ABC  and  XW  and  YZ  are  perpen- 
diculars to  AB,  then  A  BX  W  ^  hBYZ.  ^ 


w    z 

Ex.  49.     If  chords  AB  and  CD  of  a  circle  intersect  at  E  within  the 
circle,  then  A  AED  ~  A  BEC. 

Ex.  50.     If  AD  and  CE  are  the  altitudes  drawn  from  A  and  O  respec- 
tively in  A  ABC,  then  AABD-^A  CBE. 

Ex.  51.     In  the  figure  for  Ex.  50,  if  AD  intersects  CE  at  0,  prove 
A  ^£"0  is  similar  to  A  CDO. 

Ex.  52.     If  a  line  be  drawn  parallel  to  the  base  of  a  triangle  intersect- 
ing the  other  two  sides,  the  triangle  formed  is  similar  to  the  given  triangle. 

278.  Homologous  sides  of  similar  triangles  are  proportional. 

Homologous  sides  lie  opposite  equal  angles.  The  following 
exercise  illustrates  a  device  for  selecting  the  homologous  sides  and 
for  forming  the  three  equal  ratios. 


SIMILAR   POLYGONS 


157 


Illustrative  Exercise 

Hypothesis.     In  A  ABC: 

AE±BC', 
BD±AC. 


n      1     .         BD     BC 

Conclusion.    — —  =  -— 

AE     AC 


Proof. 


CD 
EC 

1.    InAAECsLTidABDC: 
Z3  =  Z4;  ZC=ZG', 


2. 

.-.  A  AEC  -  A  BDC, 

3. 

In  A  AECsiud  A  BDC: 

AC 

Z3  =  Z4 

BC 

AE 

Z(7=Z(7 

BD 

EC 

Z2  =  Z1 

DC 

A 

AC     AE 

EC 

Why? 
Why? 


Read 

Note  1 

now. 


4. 


BC     BQ     DC 


§278 


Note  1.  —  (a)  Below  A  AEC^  write  its  three  angles  and  to  the  left  of 
them  write  the  sides  which  are  opposite  them  in  A  AEC. 

(6)  Opposite  the  angles  of  A  AEC  write  the  equal  angles  of  A  BDC. 

(c)  Beside  the  angles  of  A  BDC,  write  the  sides  opposite  them  in 
A  BDC. 

(d)  Then  ^Cand  BC  are  horaologoua  sides  ;  also  AE  and  BD',  also 
EC  and  DC. 

Note  2.  -^  Sometimes  only  two  equal  ratios  are  wanted.     In  that  case, 
cross  out  in  step  4  the  ratio  that  is  not  wanted. 

279.   Principle  IV.     Four  segments  can  be  proved  propor- 
tional by  proving  them  homologous  sides  of  similar  triangles. 

Ex.  53.    If  ^5  ±^C  and  DC  ±50,  prove  that  Ap^^ 

AB^BO^AO^  J^"-"--^o       c 

DC     CO     do'  ^d 

Ex.  54.     If  the  altitudes  AD  and  CE  oiAABC  intersect  at  F,  prove 
that^F:  CF=EF'.DF. 

Ex.  55.     In  the  figure  for  Ex.  54,  prove  that  AD  :  CE  =  AB  :  CB. 
Ex.  56.     Prove  that  the  diagonals  of  a  trapezoid  divide  each  other 
so  that  the  corresponding  segments  are  proportional. 

Note.  —  Supplementary  Exercises  6  to  13,  p.  290,  can  be  studied  now. 


158 


PLANE    GEOMETRY  — BOOK   III 


Proposition  VI.     Theorem 

280.  Two  triangles  are  similar  if  an  ajigle  of  one 
equals  an  ayigle  of  the  other  and  the  sides  including 
these  angles  are  proportional. 


Hypothesis.         In  A  ABC  and  A  DEF : 

'     DE     DF 


Conclusion. 


A  ABC -^  A  DEF. 


Proof.     1.   Place  A  DEF  in  the  position  AXY,  Z  D  coincid 
ing  with  its  equal  Z.  A,  E  falling  at  X  and  F  at  Y. 
AB^AC 
AX     AY 
.-.  XY  II  BO. 
.:  A  AXY ^  A  ABC. 
.'.  A  DEF '^  A  ABC. 


2.    Then 

3. 
4. 

5. 


Hyp.  and  step  1 


Why? 

Give  the  full  proof. 

Why? 


Ex.  57.  Two  segments  AOB  and  COD  intersect  at  0  so  that  AG — 
3  OB  and  CO  =  S  CD.     Prove  AC  =  8  BD. 

Ex.  58.  ZAoi  A  ABC  is  a  right  angle.  From  E,  any  point  of  AC^ 
ED  is  drawn  perpendicular  to  BC^  meeting  it  at  Z>.  (a)  Examine  the 
figure  to  discover  a  pair  of  similar  triangles  ;  (&)  prove  the  triangles 
similar ;  (c)  from  these  triangles  determine  the  three  equal  ratios  of 
sides  of  the  triangle. 

Ex.  59.  Z  ABC  is  an  acute  angle.  CD  is  perpendicular  to  AB 
and  AF  is  perpendicular  to  BC.  (a)  Discover  a  pair  of  similar  tri- 
angles ;  (&)  prove  the  triangles  similar  ;  (c)  write  down  three  equal 
ratios  of  sides  of  these  triangles. 

Ex.  60.  The  shadow  of  a  chimney  is  36  yd.  long.  At  the  same  time 
the  shadow  of  a  stake  2  yd.  long  is  1.5  yd.  in  length.  How  high  is  the 
chimney  ? 


SIMILAR   POLYGONS 


159 


Proposition  VII.     Theorem 

281.  Two  triangles  are  similar  if  their  homologous 
sides  are  proportional. 


Hypothesis 


Conclusion. 
Proof.     1. 
Draw  XY. 

2.   Then 

3. 
4. 

5. 

6.  But 


In  A  ABC  and  A  DEF: 

AB^AC^BC 

DE     DF     EF 

A  ABC '^  A  DEF. 

On  AB,  take  AX  =  DE :  on  AC  take  A  Y=  DF. 


AB 
AX 

.'.  XT 
.'.  AAXY 
AB^BC 
AX 

AB 
DE 


,  or  

XY        DE 


AC 
AY 
BC. 

A  ABC     Give  the 
AB^BC 

xy' 

BC 

ef' 


~T^  =  -^p^-  By  hyp.  and  step  1 

Why? 
full  proof. 


Why? 
Why? 


Then 


.-.  XY=EF. 
AAXY^ADEF. 
.  A  DEF  ^  A  ABC. 


Note.  —  Notice  that  A  DEF  is  not  superposed  on  A  ABC ; 
A  AXY  is  constructed,  and  is  proved  similar  to  A  ABC  and 
A  DEF. 


§251 

Give  the  full  proof. 

Why? 

that,  rather, 
congruent  to 


Ex.  61.  Construct  any  scalene  triangle.  Then  construct  a  triangle 
whose  sides  are  double  the  corresponding  sides  of  the  first  triangle.  Are 
the  two  triangles  similar  ? 

Ex.  62.  Determine  three  segments  which  shall  bear  to  the  sides  of  a 
given  triangle  the  ratio  3:2.  Then  construct  the  triangle  having  the  new 
segments  as  sides.     Are  the  two  triangles  similar  ? 


160 


PLANE    GEOMETRY  —  BOOK   III 


Proposition  YIII.     Theorem 

282.   Homologous  altitudes  of  similar  triangles  have 
the  same  ratio  as  any  two  homologous  sides, 

4  X 


Hypothesis.  AABC^/^XYZ. 

AD  and  XTTare  homologous  altitudes. 

Conclusion.  ^  =  ^=^  =  ^. 

XW     XY      YZ     XZ 

Proof.     1.    In  rt.  A  ABD  and  rt.  A  XYW: 
2.  .'./\ABD^/\XYW. 

"  XW     XY 

4.  But^  =  ^  =  ^. 

XY      YZ      XZ 

g  .    AD  ^AB  ^BO^AG 

YZ  ' 


XW     XY 


XZ 


§272 
Why? 

Why? 

Why? 

Ax.  1,  §  51 


Note. — It  can  be  proved  that  any  two  homologous  lines  of  similar 
triangles  are  proportional  to  any  two  homologous  sides. 

Ex.  63.  The  base  and  altitude  of  a  triangle  are  5  ft.  and  3  ft.  re- 
spectively. If  the  homologous  base  of  a  similar  triangle  is  7  ft.,  find  its 
homologous  altitude. 

Ex.  64.  Prove  that  the  bisectors  of  homologous  angles  of  similar  tri- 
angles have  the  same  ratio  as  any  two  homologous  sides. 

Suggestion.  —  The  length  of  the  bisector  is  the  length  of  the  segment  of 
the  bisector  between  the  vertex  of  the  angle  and  the  opposite  side  of  the 
triangle. 

Ex.  65.  Prove  that  two  homologous  medians  of  two  similar  triangles 
have  the  same  ratio  as  any  two  homologous  sides  of  the  triangles. 

Suggestion.  —  Use  §  280. 


SIMILAR   POLYGONS  161 

Proposition  IX.     Theorem 

283.  If  tivo  chords  are  draion  through  a  fixed  point 
within  a  circle,  the  product  of  the  segments  of  one  is 
equal  to  the  product  of  the  segments  of  the  other. 


Hypothesis.  AB  and  CD  are  any  two  chords  of  O  0  inter- 
secting at  point  P. 

Conclusion.  APPB^DP-  PC. 

Analysis.     1.   ltAP'PB  =  DP-  PC,  then  AP :  PC  =  DP  :  PB. 

§252 
2.    .-.  try  to  prove  A  APD  ~  A  PBC.  §  279 

Proof.     1.  Draw  AD  and  BC. 

2.  A  APD  ~  A  PBC.  Give  the  full  proof. 

3.  ...^=^.  Why? 

PC      PB  ^ 

4.  .-.  AP'PB  =  PC'  DP.  Why  ? 

284.  Principle  V.  To  prove  that  the  product  of  two  seg- 
ments equals  the  product  of  two  other  segments,  first  derive 
from  the  equation  a  proportion  by  §  252  and  then  try  to  pro- 
ceed as  in  §  279. 

This  principle  is  illustrated  in  the  proof  of  §  283. 

Ex.  66.  Two  chords  of  a  circle  intersect  so  that  the  segments  of  one 
are  4  in.  and  5  in.  respectively.  If  the  shorter  segment  of  the  other  is 
6  in.,  what  is  the  longer  segment  ? 

Ex.  67.  In  a  circle  whose  diameter  is  16  in.,  a  chord  14  in.  long  is 
drawn  through  a  point  which  is  4  in,  from  the  center.  What  are  the  two 
segments  of  the  chord  ?     (Represent  one  segment  by  x.) 


162 


PLANE    GEOMETRY  —  BOOK   III 


285.  If  a  secant  PA  is  drawn  to  a  circle 
from  a  point  P,  cutting  the  circle  at  point 
B,  then  PA  is  called  the  whole  secant,  PB 
the  external  segment,  and  AB  the  internal 
segment. 

Propositio:n"  X.     Theorem 

286.  If  any  two  secants  are  drawn  through  a  fixed 
point  outside  a  circle^  the  product  of  one  and  its  ex- 
ternal segment  equals  the  product  of  the  other  and  its 
external  segment. 


Hypothesis.     ABP  and  CDP  are  two  secants  of  O  0. 
Conclusion.  AP'BP=  CP-  DP. 

Suggestion.  — Make  an  analysis  and  a  proof  similar  to  that  for  Proposition 
IX. 

Ex.  68.  From  a  point  P,  a  secant  18  in.  long  is  drawn  to  a  circle  ; 
the  external  segment  is  3  in.  The  external  segment  of  a  second  secant 
from  the  same  point  is  6  in.  long.     How  long  is  the  whole  secant  ? 

Ex.  69.  A  secant  is  drawn  from  point  P  to  a  circle.  The  external 
segment  is  4  in.  and  the  internal  segment  is  6  in.  How  long  must  a 
second  secant  be  in  order  that  its  internal  segment  shall  be  3  in.  ? 

Ex.  70.  If  from  an  exterior  point  P,  any  number  of  secants  be 
drawn,  the  product  of  the  whole  secant  and  the  external  segment  is 
constant. 

Note.  —  The  conclusion  means  that  the  product  of  the  whole  secant 
and  the  external  segment  is  the  same  for  each  secant. 

Ex.  71.  Prove  that  the  product  of  the  segments  of  one  diagonal  of 
an  inscribed  quadrilateral  is  equal  to  the  product  of  the  segments  of  the 
other  diagonal. 


SIMILAR   POLYGONS  163 

Proposition  XI.     Theorem 

287.  If  a  secant  and  a  tangent  are  draioii  to  a  circle 
from  the  same  point  outside  a  circle,  the  square  of  the 
tangent  is  equal  to  the  product  of  the  whole  secant  and 
its  external  segment. 


H3rpothesis.     PC  is  a  tangent  to  O  0 ;  secant  PA  intersects 
the  O  at  ^  and  A. 
Conclusion.  CP"^  =  AP  -  BP. 

(Analysis  and  proof  left  to  the  pupil.) 
Note.  —  Proposition  XI  maybe  stated:  If  a  secant  and  a  tangent  are 
drawn  to  a  circle  from  the  same  point  outside  the  circle,  the  tangent 
is  the  mean  proportional  between  the  whole  secant  and  its  external 
segment. 

For,  when        CP  =  AP  ■  BP,  AP  :  CP  =  CP :  BP.  Why  ? 

Ex.  72.  The  length  of  the  tangent  to  a  circle  from  a  point  outside 
is  12  in.  What  must  be  the  length  of  a  secant  from  the  same  point  in 
order  that  the  external  segment  will  be  8  in.  ? 

Ex.  73.  If  altitudes  AD  and  BE  of  A  ABC  intersect  at  F,  prove 
that  the  product  of  the  segments  of  one  is  equal  to  the  product  of  the 
segments  of  the  other.     (Apply  §  284.) 

Ex.  74.  If  AD  and  BE  are  two  altitudes  of  A  ABC,  then  AD  ■  BC 
=  BE.AC. 

Ex.  75.  If  Z  J.  of  A  ABC  is  a  right  angle  and  ED  is  drawn  perpen- 
dicular to  CB  from  any  point  E  of  AB,  meeting  CB  at  D,  then  EB  ■  AB 
=  CB .  DB. 

Ex.  76.   If  altitudes  AD  and  BE  of  A  ABC  intersect  at  F,  then  : 
(a)  BE  •  EF=  AE'EC;  (6)  AD  ■  DF  =  BD  •  DC. 

Note.  —  Supplementary  Exercises  14  to  18,  p.  290,  can  be  studied  now. 


164 


PLANE    GEOMETRY  —  BOOK   III 


Proposition^  XII.     Theorem 

288.  If  the  altitude  he  draivn  to  the  hypotenuse  of  a 
right  triayigle : 

I.  The  altitude  is  the  mean  proportional  hetiveen  the 
segments  of  the  hypotenuse  ; 

II.  Each  leg  is  the  mean  propo7'tional  hetioeen   the 
whole  hypoteiiuse  and  the  adjacent  segment. 


Hypothesis.     In  A  ABC,  Z  (7  is  a  rt.  Z. 
CD  ±  AB, 

Conclusion.    I.  ^^  =  ^. 

CD     DB 

Proof.     1.    Z  3  and  Z  1  are  each  complements  of  Z  2. 

Why? 

.-.  Z  1  =  Z  3.  Why  ? 

2.  .-.  A  ADC  ~  A  DCB,       Give  the  full  proof. 

"CD    db' 


See  §  278 


Conclusion.     II.      (a) 

Plan.     For  («)  prove  A  ABC '^  A  ACD. 


AB^AC,   ...    AB^BC 
AC     AD'  ^  ^   BC     db' 


Cor.     If  a  perpendicular  be  drawn 
from  any  point  on  a  circle  to  a  diameter  : 

(a)  The  perpendicular  is  the  mean  propor- 
tional between  the  segments  of  the  diameter. 

(b)  The  chord  joining  the  point  to  either 
extremity  of  the  diameter  is  the  meayi  proportional  between  the 
whole  diameter  and  the  segment  of  it  adjacent  to  the  chord. 


SIMILAR   POLYGONS  165 

Hx.  77.  Find  the  altitude  drawn  to  the  hypotenuse  of  a  right  tri- 
angle if  it  divides  the  hypotenuse  into  two  segments  whose  lengths  are 
3  in.  and  12  in.  respectively.    Find  each  leg  of  the  riglit  triangle. 

Ex.  78.  The  hypotenuse  of  a  right  triangle  is  20  in.  and  the  perpen- 
dicular to  it  from  the  opposite  vertex  is  8  in.  Find  the  segments  of  the 
hypotenuse,  and  the  two  legs  of  the  triangle. 

Ex.  79.     C  and  D  are  respectively  the  mid-points  of  a  chord  AB  and  its 
subtended  arc.     If  J.i>  =  12  and  CZ>=8,  what  is  the  diameter  of  the  circle  ? 
Suggestion.  —  DC  extended  passes  through  the  center  of  the  circle. 

Ex.  80.  A  chord  of  a  circle  is  20  in.  in  length.  Its  mid-point  is  5  in. 
from  the  mid-point  of  its  arc.    Find  the  diameter  of  the  circle. 


Proposition  XIII.     Theorem 

290.   Construct  the   mean  proportional  between  two 
given  segments.  j) 


A'- 


\ 

1 

B''  C 


^.4_n_-...l_._._E 


Given  segments  m  and  n. 

Required  to  construct  the  mean  proportional  between  m  and  n. 

Construction.     1.   On  line  AE  take  AB  =  m  and  BC  =  n. 

2.  Construct  semicircle  ADC  on  AC  as  diameter. 

3.  Construct  DB  _L  AC,  meeting  the  semicircle  at  D. 
Statement.     DB  is  the  mean  proportional  between  m  and  n. 

(The  proof  Ls  to  be  given  by  the  pupil.     See  §  289.) 

Ex.  81.    Construct  a  segment  equal  to  aV3  where  a  is  any  segment 
whatever. 

Analysis.     1.   Let  a;  =  a  V3.    Then  x^  =  3  a^.  Why  ? 

2.  .-.  3a  :  X  =a;:a.  §  262 

3.  .-.  X  is  the  mean  proportional  between  a  and  3  a. 

(Construction  is  to  be  given  by  the  pupil.) 

Ex.  82.     Construct  a  segment  equal  to  VZab  where  a  and  b  are  any 
given  segments. 


166  PLANE    GEOMETRY  —  BOOK   III 

•   Proposition  XIV.     Theorem 

291.    The  square  of  the  hypotenuse  of  a  right  triangle 
is  equal  to  the  sum  of  the  squares  of  the  legs. 


Hypothesis.     In  A  ABC,  Z  (7  is  a  right  angle. 

Conclusion.  c^  =  a^  +  b\ 

Proof.     1 .    Draw  CD  ±  AB.     Let  AD  =  r,  and  DB  =  s. 


2. 

Then 

as          h     r 

§  288,  II 

3. 

.'.  a^  =  cs  and  ¥  =  cr. 

Why? 

4. 
5. 

.-.  a^  -\- b"^  =  cs -\- cr 

=  c(s-^r). 
...  ct2  +  62  =  c  •  c,  or  a2  +  62  ^  c^. 

Note.  —  This  theorem  is  called  the  Pythagorean  Theorem,  after  Pythag- 
oras, who  formulated  it.  The  theorem  was  evidently  known  even  to  the 
Egyptians.  This  proof  of  the  theorem  is  attributed  to  Hindu  mathema- 
ticians. In  Book  IV,  we  shall  study  Euclid's  proof  of  the  theorem  —  a 
strictly  geometric  proof,  whereas  this  is  more  an  algebraic  one. 

292.  Cor.  TJie  square  of  either  leg  of  a  right  triangle  is 
equal  to  the  square  of  the  hypotenuse  minus  the  square  of  the  other 
leg. 

Ex.  83.  How  long  must  a  rope  be  to  run  from  the  top  of  a  12-foot 
tent  pole  to  a  point  16  ft.  from  the  foot  of  the  pole  ? 

Ex.  84.  The  diameters  of  two  concentric  circles  are  14  in.  and  60  in., 
respectively.  Find  the  length  of  a  chord  of  the  greater  circle  which  is  tan- 
gent to  the  smaller. 

Ex.  85.  A  baseball  diamond  is  a  square  whose  sides  are  each  90  ft. 
Ions;.     What  is  the  distance  from  "  first"  to  "  third"  ? 


SIMILAR   POLYGONS  167 

Ex.  86.     Find  the  formula  for  the  diagonal  of  a  square  whose  side  is  s. 
By  this  formula  determine  the  diagonal  when  :    (a)  s  =  10  ;  (b)  s  =  15. 

Ex.  87.  The  equal  sides  of  an  isosceles  trapezoid  are  each  10  in.  long. 
One  of  the  bases  is  30  in.,  and  the  other  is  42  in.  in  length.  What  is  the 
altitude  of  the  trapezoid  ? 

Ex.  88.  Find  the  length  of  the  altitude  of  an  equilateral  triangle  if 
each  side  is  10  in. 


Ex.  89.    Derive  the  formula  for  the  length  of  the  alti- 
tude of  an  equilateral  triangle  if  each  side  of  the  triangle  is  s. 
By  this  formula  determine  the  altitude  when  :   ,  ^ 

{a)  s  =  6in.  ;  (6)  8  =  13  in. 

Ex.  90.     A  piece  of  silk  27  in.  wide  is  folded  "  on  the 
bias  "  along  the  line  AB.     How  long  is  AB  ? 

Ex.  91.  Find  the  length  of  each  side  of  a  rhombus  if  the  diagonals 
are  6  in.  and  8  in.  respectively. 

Ex.  92.     If  AD  is  the  perpendicular  from  A  to  BC  of  A  ABC,  prove 

AB^  -  'AO^  =  DB^  -  CD^. 

Plan.  Find  an  expression  for  Alf  and  AC^ ;  then  subtract  the  latter 
from  the  former. 

Note.  —  This  might  be  called  a  "common  sense"  plan.  After  form- 
ing in  this  manner  the  left  member,  if  the  right  member  is  not  obtained  at 
once,  then  proceed  to  form  in  the  same  manner  the  right  member,  and 
afterwards  try  to  prove  the  two  values  obtained  are  equal. 

Ex.  93.  If  D  is  any  point  in  the  altitude  from  A  to  side  BC  ot 
A  ABC,  prove  that  A^  -  AC^  =  ~DB^  -  DC\ 

Suggestion.  —  Read  the  note  under  Ex.  92. 

Ex.  94.  If  a  parallel  to  hypotenuse  AB  of  right  triangle  ABC 
meets  AC  and  BC  at  D  and  E  respectively,  prove  that 

'AE'^  4-  BD^  ='AB^  +  DE^. 

Ex.  95.     If  perpendiculars  PF,  PD,  and  PE  be 

drawn  from  any  point  P  within  an  acute-angled  tri- 
angle ABC  to  sides  AB,  BC,  and  CA  respectively,       F 
prove  that 

AF^  -^  BB^  -\-CE^  =  AE^  +^F^  +  CJD^.  "  D 

Note.  — Supplementary  Exercises  19  to  40,  p.  291,  can  be  studied  now. 


168 


PLANE    GEOMETRY  —  BOOK   III 


Proposition   XY.     Theorem 

293.  Two  polygons  are  similar  if  they  are  composed 
of  the  same  number  of  triangles,  similar  each  to  each, 
and  similarly  placed. 


Hypothesis.     A  AEB  ^  A  A'E'B' ;  A  EBD  ~  A  E'B'D' ; 
ABCD^AB'C'D'. 
The  triangles  are  similarly  placed. 

Conclusion.     Polygon  ABODE  ~  polygon  A'B'C'D'E'. 

Analysis.     The  homologous  A  must  he  proved  equal,  and  the  homolo- 
gous sides  must  he  proved  proportional.  §  272 

Proof.     1.         Z1  =  Z1';     Z2  =  Z2';     Z3  =  Z3'.  Why? 

2.  .-.  ZjB  =  Z5'.  Why? 

3.  Similarly,  Z  Z>  =  Z  Z)' ;  ZE  =  ZE'',  ZA^ZA'; 

z  c=za. 

BE 


4. 
5. 

6.  Similarly, 

7.  Also 
8. 


AE_^BE^    ^^^   ED^^ 

A'E'     B'E''  E'D'     B'E' 

AE       ED 


Prove  it. 
Why? 


A'E'  E'D' 
ED  ^  CD 
E'D'      CD' ' 

^  =  ^,and^ 
A'B'     A'E"  B'C 


CD 


AB 


AE       ED        CD 


CD' 

^ ^ ^ ^  BC 

A'B'     A'E'     E'D'      CD'     B'C' 
9.   .-.  Polygon  ABCDE  ^  polygon  A'B' CD' E'. 


Ax.  1,  §  51 

Why? 

Ax.  1,  §  51 
Why? 


SIMILAR  POLYGONS 


169 


Proposition  XVI.     Problem 

294.  Upon  a  given  segment ^  homologous  to  a  given 
side  of  a  given  polygon,  construct  a  polygon  similar  to 
the  given  polygon. 


Given  polygon  ABCDE  and  segment  A'B'. 

Required  to  construct  upon  A'B'  as  side  homologous  to  AB 
a  polygon  similar  to  ABCDE. 

Construction.     1.    Divide   ABCDE  into  triangles  by  draw- 
ing EB  and  BD. 

2.  Construct  A  A'B'E'  similar  to  A  ABE,  by  making  Z  A' 
=  ZA,  andZl'  =  Zl. 

3.  Construct  A  E'B'D'  similar  to  A  EBD.  How  ? 

4.  Construct  A  D'B'C  similar  to  A  DBC.  ■  How  ? 
Statement,     polygon  A'B'CD'E'  ~  polygon  ABCDE. 

Why? 
Ex.  96.  ABCD  is  the  shape  of  an  irregular 
piece  of  groimd.  Make  a  figure  similar  to  ABCD 
such  that  each  side  of  the  resulting  figure  shall  be 
three  times  as  long  as  the  corresponding  side  of  the 
given  figure. 

Ex.  97.    An  ordinary  shed  roof  is  said  to  have  a 
the  distance  AB  is  one  third  of  the  distance  CD. 

A  carpenter  wishes  to  order  some  "  two  by  fours  ""  for 
the  rafters  AE  of  a  garage  which  is  to  be  24  ft.  wide  and 
have  a  one-third  pitch.  He  allows  \  ft.  for  cutting  at 
the  end  A^  and  wants  the  rafters  to  project  beyond  the 
wall  at  C  so  that  CE  will  be  2  ft.  What  length  of  "  two 
by  fours  "  must  he  order  if  they  can  be  obtained  only  in  even  lengths  ? 


one-third  pitch 


170 


PLANE    GEOMETRY  —  BOOK   III 


Pkoposition  XVII.     Theorem 

295.  Tivo  similar  polygons  can  he  divided  into  the 
same  number  of  triangles,  similar  each  to  each,  and 
similarly  placed. 


Hypothesis.  Polygon  ABODE  ~  polygon  A'B'C'D'E',  ho- 
mologous vertices  being  indicated  by  corresponding  letters. 

Conclusion.  The  polygons  can  be  divided  into  the  same 
number  of  triangles,  similar  each  to  each  and  similarly 
placed. 

Construction.     Draw  BE,  BD,  B'E',  and  B'D'. 
Statement.     A  ABE  ~  A  A'B'E' ;  A  EBD  ~  A  E'B'D' ; 
A  BCD  ~  A  B'C'D'. 

AB       AE 


Proof.     1. 

2. 
3. 


4. 


ZJ.  =  Zyl'and  ,        ^,    , 

A'B'     A'E' 

.-.  A  BAE  ~  A  B'A'E'. 

ZE  =  ZE'  Sind  Z  4:  =  Z  4'. 

.-.  Z5  =  Z5'. 

BE       AE         -,    T       ED       AE 


B'E' 


AE         .    1       ED 

— ; — -,  and  also  -^ 

A'E''  E'D' 


A'E' 


BE 


ED 


5. 

6.    Similarly 


B'E'     E'D' 
.'.  A  EBD  ~  A  E'B'D'. 
A  BCD  ~  A  B'C'D'. 


Why? 

Why? 
Why? 
Why? 

Why? 

Why? 
Why? 


Note.  —  If  X  and  T  are  any  two  points  of  one  polygon  and  X'  and 
Y'  are  the  homologous  points  of  a  similar  polygon,  then  XZand  XT' 
are  homologous  segments  and  XY :  X'Y'  equals  the  ratio  of  similitude. 


SIMILAR  POLYGONS  l7l 

296.  Fundamental  Theorem  about  Equal  Ratios. 

In  a  series  of  equal  ratios^  the  sum  of  the  antecedents  is  to  the 
sum  of  the  consequents  as  any  antecedent  is  to  its  consequent. 

If  -=-  =  -  =  1 

b     d     f     h' 

then  ^±_^+l+^  =  «  =  ^  =  etc. 

h-{-d-\-f-^h     b      d 

Proof.     1.   Let  r  =  -  and  hence      br  =  a. 
b 

2.  .-.  dr  =  c,  fr  =  e,  hr  =  g.  Prove  it. 

3.  .'.  br -\- dr -\- fr -^  hr  =  a -\- c -^  e  +  g.  Why? 

4.  ...,.=:!L±A+i+Lg.  Prove  it. 

b  +  d+f-hh 

5.  Hence  ^Jl^_±l+5  =  «  =  £  =  etc. 

b  +  d-{-f-{-h     b      d 

Proposition  XVIII .     Theorem 

297.  JTie  perimeters  of  two  similar  polygons  have  the 
same  ratio  as  any  tioo  homologous  sides. 

B 

b' 


D  E  U 

Hypothesis.     ABCDE  and  A'B'C'D'E'  are  similar  polygons 
with  homologous  vertices  indicated  by  corresponding  letters. 

Conclusion. 

AB  +  BC+CD  +  DE-^EA      ^  AB  ^  BO  ^ 
A'B'  +  B'C  +  CD'  +  D'E'  +  E'A'     A'B'     B'C     ^  ^' 

Proof.     1.  -^  =  ^  =  -^  =  etc. 

A'B'     B'C     CD' 

(Complete  the  proof.     Apply  §296.) 
Note.  —Supplementary  Exercises  41  to  44,  p.  293,  can  be  studied  now. 


172  PLANE    GEOMETRY  — BOOK   III 

SUPPLEMENTARY  TOPICS 

Five  groups  of  theorems  follow,  —  all  of  which  appear  in 
modern  geometries.  It  is  not  necessary, — in  fact,  it  may  be 
unwise  to  study  all  of  them  in  every  class.  The  teacher 
should  feel  free  to  select  the  group  or  groups  which  appear  to 
be  of  most  value  to  the  class. 

Each  group  is  independent  of  each  of  the  others. 

None  of  these  theorems  is  required  as  an  authority  in  the  main 
lists  of  theorems  of -succeeding  Books. 

Group  A,  —  Scales  and  Scale  Drawing. 

Group  B.  —  Trigonometric  Ratios  and  their  Application. 

These  two  groups  are  interesting  and  valuable  applications  of 
Book  III. 

Group  C.  —  Proportional  Division  of  a  Segment. 
Group  D.  —  External  and  Harmonic  Division  of  a  Segment. 
Group  ^. — Numerical    Relations    among   Segments    of    a 
Triangle. 

These  last  three  groups  have  long  appeared  in  geometries. 

A.   Scales  and  Scale  Drawing 

298.  Scale  drawings  are  a  common  and 
useful  application  of  similar  polygons. 

The  adjoining  figure  represents  a  lot  150'  x 
276'.  It  is  drawn  to  the  scale  of  1"  to  200  ' ;  that 
is,  AB,  I"  in  length,  represents  150'  and  jBC,  If"  in 
length,  represents  275'.  If  the  corners  of  the  lot 
itself  are  denoted  by  J.',  B',  C,  and  D'  respectively, 
then  AB  :  A'B'  =-.  1 :  2400,  and  5(7 :  5'  C  =  1  :  2400. 

(1"  to  200'  is  1"  to  2400"  or  1 :  2400.) 

ABCD  is  similar  to  A'B' CD',  for  the  two  figures 
are   mutually  equiangular  (being  rectangles)    and 
their  homologous  sides   are  proportional  (the  ratio  of  similitude  being 
1:2400). 

Since,  in  similar  figures,  the  ratio  of  any  two  homologous  sides  equals 
the  ratio  of  similitude  (Note,  §  295),  it  is  possible  to  determine  from 
ABCD  the  approximate  length  of  any  segment  on  the  field  itself. 


SCALES  AND   SCALE   DRAWING  173 

299.  Scales.  The  construction  and  use  of  scale  drawings 
are  made  easy  by  the  construction  in  advance  of  the  scale  it- 
self, unless  such  a  scale  is  already  at  hand. 

Example.  —  Below  is  the  scale  of  1"  to  100'  to  measure 
350  ft. 

Scale   r  =    100 

The  segment  extending  from  the  zero  mark  to  any  division 
point  represents  the  number  of  feet  indicated  above  that  point. 
Notice  that  the  left-hand  section  is  divided  into  ten  equal 
parts. 

To  determine  the  number  of  feet  represented  by  a  given 
segment  according  to  the  given  scale :  take  the  segment  on  the 
dividers ;  place  one  point  of  the  dividers  on  a  division  mark 
at  the  right  of  the  zero  mark,  so  that  the  other  point  of  the 
dividers  will  fall  on  the  scale  either  at  the  zero  mark  or  to  the 
left  of  it.  The  length  represented  by  the  segment  may  then 
be  read  to  the  nearest  5  feet. 

Thus,  the  segment  a  below  represents  235  ft.  if  drawn  to  the  scale  of 
1"  to  100'. 


Ex.  98.     Determine  the  length  represented  by  each  of  the  following 
segments,  assuming  that  they  are  drawn  to  the  scale  of  1"  to  100'. 


Ex.  99.  Determine  the  approximate  number  of  feet  represented  by 
the  diagonal  ^C  in  the  figure  of  §  298. 

Ex.  100.  Determine  the  approximate  distance  of  the  tree,  T,  from 
each  of  the  comers  of  the  lot  ABCD  in  the  figure  of  §  298. 

Ex.  101.  Construct  the  scale  of  1"  to  1'  to  measure  5  ft.,  having 
the  left-hand  section  show  the  segments  corresponding  to  2",  4",  etc., 
to  12". 

What  length,  in  feet  and  inches,  do  segments  a,  6,  and  c  of  Ex.  98 
represent  if  it  is  assumed  that  they  are  drawn  to  the  scale  of  1"  to  1'  ? 


174 


PLANE    GEOMETRY  — BOOK   III 


Ex.  102.  Construct  the  scale  of  1"  to  25',  to  measure  100  ft.,  having 
the  left-hand  section  show  segments  corresponding  to  5',  10',  etc.,  to  25', 

What  length  do  segments  a,  6,  c  of  Ex.  98  represent  if  it  is  assumed 
that  they  are  drawn  to  the  scale  of  1"  to  25'  ? 


Ex.  103.  Draw  to  the  scale  of  1"  to  25'  the  ad- 
joining figure.  From  the  figure  so  drawn,  determine 
the  approximate  height  of  AB. 


Ex.  104.  Draw  to  the  scale  of  1"  to  25'  a 
figure  similar  to  the  adjoining  one.  From  the 
resulting  figure,  determine  the  approximate  dis- 
tance represented  by  AB. 

Ex.  105.  Draw  to  the  scale  of  1"  to  25'  a  figure 
similar  to  the  adjoining  one.  From  the  resulting 
figure,  determine  the  approximate  length  of  AB,  if 
CD  =60',  ZACB  =  QQ°,  ZADC=ZO'',  said  Z  ABC 
=  90°. 


Ex.  106.     The  perimeters  of  two  similar  polygons  are  119  and  68  ;  if  a 
side  of  the  first  is  21,  what  is  the  homologous  side  of  the  second  ?    (§  297.) 

Ex.  107.  Draw  to  the  scale  of  1"  to  100'  a 
figure  similar  to  the  adjoining  one.  From  the 
resulting  figure,  determine  the  approximate  perime- 
ter of  the  field  having  the  dimensions  indicated. 


B.   Trigonometric   Ratios   and   their 
Application 

300.   Sine  of  an  Angle.     Let  Z  ABO  be  any  angle, 
on  BC  any  points  Pi  and  Pg-     Draw  per- 
pendiculars PiEi   and  P2R2  to  AB. 

Then      ABP^R.^ABPA-     (Why?) 

.   R\P\ -K2P2 

**  'BPi~'BP,' 

That  is,  the  ratio  of  the  perpendicular  BP  to  the  distance 
BP  is  the  same,  regardless  of  where  P^  and  P^  are  located 
onjBC 


TRIGONOMETRIC  RATIOS 


175 


This  constant  ratio  is  called  the  Sine  of  Z  B.     (Sin  B.) 
When  the  angle  is  acute,  its  sides  and  the  perpendicular 
form  a  right  triangle.     In  this  triangle, 

sine  of  acute  angle  =  side  opposite  -f-  hypotenuse. 

The  sine  of  a  given  angle  may  be  computed  as  in  the  fol- 
lowing 

Example.  —  Let  Z  B=  60°.     Determine  sin  60°. 

Solution.  1.    Draw  Pi?  ±^A   DrawPT^P^. 

2.  Then  APBT^A  PBB,  and  Z  T  =  60°. 

(Why?) 

3.  .-.  A  PP7'is  equilateral,  and  BB  =  ^  BP. 

(Why  ?) 

4.  Let  BP  =  2  m  and  hence  BB  =  m. 

5.  In  A  BPB,  BP^  =  4  m^-  m2  =  3  m2._  (Why  ?) 

6.  .-.  i?P  =  »nV3. 


...sin  60°  =  ^=??^  =1^232 
BP       2  m  2 


.866+. 


Ex.  108.     Determine  as  in  the  example  the  value  of  sin  45°  and  of 

sin  30°. 

Ex.  109.  Construct  a  figure  similar  to  the  ad- 
joining one  making  :  /.  ABC  =  35°  ;  Z  ABD  =  50°  ; 
Z  ABE  =  65°  ;  and  Z  ABF  =  75°.  Draw  the  per- 
pendiculars from  C,  D,  E,  and  F  to  AB.  Measure 
these  perpendiculars  and  also  the  radius.  Then  com- 
pute the  approximate  values  of  the  sine  of  each  of 
the  angles  indicated  ;  that  is,  of  sin  35°,  sin  50°, 
sin  65°,  and  sin  75°. 

(If  you  have  a  metric  scale,  make  AB  =  100  mm.  and  measure  the 
perpendiculars  in  mm. ;  if  you  do  not  have  a  metric  scale,  make  AB  =  3| 
in.,  and  measure  the  perpendiculars  in  sixteentlis  of  an  inch.  If  you  use 
metric  measures,  your  sines  should  be  approximately  correct  to  the 
second  decimal  place  ;  if  you  use  the  English  scale,  the  values  should 
be  correct  to  the  first  decimal  place.  Keep  your  figure  for  use  in  a  later 
exercise.) 

Ex.  110.     Construct  the  acute  angle : 

(a)  Whose  sine  is  J  ;  (6)  whose  sine  is  f. 

Ex.  111.  BC,  40  mm.  long,  is  on  one  side  of  Z  ABC,  whose  sine  is  ^. 
How  long  is  CA,  the  perpendicular  to  side  AB  ? 


i?j       /?2 


176  PLANE    GEOMETRY— BOOK   III 

301.  Cosine  and  Tangent  of  an  Angle. 

It  is  easily  proved   that   the  ratio    -~  ^'^^ 

is  constant  for  all  positions  of  P  on  J5(7, 
as  in  §  300 ;  and  also  that is  constant.    „ 

~  is  called  the  Cosine  of  Z  B,     (Cos  5.) 

^  is  called  the  Tangent  of  Z  B.     (Tan  B^ 
BR 

In  the  right  triangle  formed  when  Z  5  is  an  acute  angle : 

cosine  of  acute  angle  =  adjacent  side  -^  hypotenuse ; 

tangent  of  acute  angle  =  opposite  side  -v-  adjacent  side. 

Example. —When  Z5  =  60°,  if  BP=2m,  then  BR  =  m 
and  EP  =  m  V3.     (See  Example,  §  300.) 

Hence,  Cos  60°  =  ^^  =  -  =  .500. 

2m      2 

Tan  60°  =V12^  =  Vs  =  1.732. 
m 

Ex.  112.  Compute  the  cosine  and  the  tangent  of  45°  and  30°  respec- 
tively. 

Ex.  113.     In  the  figure  constructed  for  Ex.  109,  measure  BM,  BN, 
BB,  and  BS.     Then  compute  the  approximate  values  of : 
(a)  cos  85°  ;  cos  50°  ;  cos  65°  ;  cos  75°  ; 
(6)   tan  35°;  tan  50°;  tan  65°;  tan  75°. 

302.  Table  of  Values  of  Trigonometric  Ratios.  The  values 
of  the  sine,  cosine,  and  tangent  of  certain  angles  have  been 
computed.     See  the  table  opposite. 

To  determine  the  sine  of  37°  from  the  table :  in  the  first 
column  find  37° ;  on  the  same  line  with  it,  and  in  the  column 
headed  by  the  word  Sin,  is  found  .602.     This  is  the  sine  of  37°. 

Similarly  the  cosine  or  tangent  of  a  given  angle  may  be 
determined  from  the  table. 


TRIGONOMETRIC  RATIOS 
Table  of  Values  of  Trigonometric  Ratios 


177 


Anolr 

Sin 

Cos 

Tan 

Angle 

Sin 

Cob 

Tan 

10^ 

.174 

.985 

.176 

45° 

.707 

.707 

1.000 

11° 

.191 

.982 

.194 

46° 

.719 

.696 

1.036 

12° 

.208 

.978 

.213 

47° 

.731 

.682 

1.072 

13° 

.226 

.974 

.231 

48° 

.743 

.609 

1.111 

14° 

.242 

.970 

.249 

49° 

.755 

.656 

1.150 

16° 

.259 

.966 

.268 

60° 

.766 

.643 

1.192 

16° 

.276 

.961 

.287 

61° 

.777 

.629 

1.236 

17° 

.292 

.956 

.306 

62° 

.788 

.616 

1.280 

18° 

.309 

.951 

.326 

63° 

.799 

.602 

1.327 

19° 

.326 

.946 

.344 

64° 

.809 

.588 

1.376 

20° 

.342 

.940 

.364 

66° 

.819 

.674 

1.428 

21° 

.358 

.934 

.384 

66° 

.829 

.659 

1.483 

22° 

.376 

.927 

.404 

67° 

.839 

.646 

1.640 

23° 

.391 

.921 

.424 

68° 

.848 

.630 

1.600 

24° 

.407 

.914 

.446 

69° 

.867 

.516 

1.664 

26° 

.423 

.906 

.466 

60° 

.866 

.600 

1.732 

26° 

.438 

.899 

.488 

61° 

.876 

.486 

1.804 

27° 

.454 

.891 

.610 

62° 

.883 

.4t)9 

1.881 

28° 

.469 

.883 

.532 

63° 

.891 

.454 

1.963 

29° 

.485 

.875 

.654 

64° 

.899 

.438 

2.050 

30^ 

.500 

.866 

.677 

66° 

.906 

.423 

2.144 

31° 

.515 

.857 

.601 

66° 

.914 

.407 

2.246 

32° 

.530 

.848 

.625 

67° 

.921 

.391 

2.356 

33° 

.545 

.839 

.649 

68° 

.927 

.376 

2.476 

34° 

.559 

.829 

.675 

69° 

.934 

.358 

2.606 

36° 

.674 

.819 

.700 

70° 

.940 

.342 

2.747 

36° 

.588 

.809 

.727 

71° 

.946 

.326 

2.904 

87° 

.602 

.799 

.754 

72° 

.961 

.309 

3.078 

38° 

.616 

.788 

.781 

73° 

.956 

.292 

3.271 

39° 

.629 

.777 

.810 

74° 

.961 

.276 

3.487 

40° 

.643 

.766 

.839 

76° 

.966 

.259 

3.732 

41° 

.656 

.756 

.869 

76° 

.970 

.242 

4.011 

42° 

.669 

.743 

.900 

77° 

.974 

.226 

4.331 

43° 

.682 

.731 

.933 

78° 

.978 

.208 

4.706 

44° 

.696 

.719 

.966 

79° 

.982 

.191 

6.146 

48° 

.707 

.707 

1.000 

80° 

.986 

.174 

6.671 

Ex.  114.     Obtain  from  the  table  :   (a)  sin  40*^ 
(b)  cos  37°  ;  cos  76°  ;  cos  64° ;  (c)  tan  29° 


sin  63°  ;  sin  26°  ; 
tan  68°  :  tan  71°. 


Ex.  115.     What  is  the  angle  x  if : 

(a)  sin  X  =  .454  ;  sin  a;  =  .829  ;  sin  x  =  .978  ; 
(6)  cos  X  =  .966 ;  cos  a;  =  .719  ;  cos  x  =  .368  ; 
(c)  tan  X  =  .424  ;  tan  x  =  1.111 ;  tan  x  =  3.732. 


178  PLANE    GEOMETRY  —  BOOK   III 

303.  Application  of  Trigonometric  Ratios. 

Example  1. — Assume  that  at  point  B 
ZCBA  =  34°  and  that  BC=125  ft.  How 
high  above  ground  is  point  A? 

Solution.      1 .   ^  =  tan  34°,  ot  AC  =  BC  xta,B.  34°.    ^ '       225^ 

2.  .-.  ^C=  125  X  .68=85ft. 

Note.  —  Z  CBA  is  called  the  Angle  of  Elevation  of  A  at  point  B. 

Example  2.  — AB  represents  a  lighthouse    j).. 
250  ft.  high.     DA  is  an  imaginary  line  paral- 
lel to  BO.     C  represents  the  position  of   a 
ship.    Z  DAC=  31°.    How  far  from  B  is  C? 


Solution. 

1.                     ZBCA  =  S1°. 

2. 

T&nSl" -"^^^  or  BC-AB  :  tan  31°. 
BC 

3. 

.•.BC=:  250  -  .601  =  415.9  ft. 

Why? 


That  is,  BCis  about  416  ft. 

Note.  — ZDACis  called  the  Angle  of  Depression  of  C at  A. 

Ex.  116.  In  the  adjoining  figure,  ii  AC ±  CD,  CD 
=  100  ft.,  and  ZD  =  62°,  what  are  the  distances  AC 
and  AD  ? 

Ex.  117.  rind  the  angle  of  elevation  of  the  sun 
when  a  monument  whose  height  is  360  ft.  casts  a  shadow 
400  ft.  in  length. 

Ex.  118.  Determine  the  length  of  one  side  of  an  equilateral  polygon 
having  nine  sides  which  is  inscribed  in  a  circle  of  radius  10  ft. 

Determine  the  distance  of  each  of  the  sides  from  the  center  of  the 
circle. 

Ex.  119.  A  boy  flying  a  kite  knows  that  he  has  500  ft.  of  twine. 
Another  boy  stations  himself  approximately  below  the  kite.  The  angle 
made  by  the  string  with  an  imaginary  line  running  from  the  first  boy  to 
the  second  is  approximately  50°.  Determine  the  approximate  height  of 
the  kite. 

Ex.  120.  At  a  point  169  ft.  from  the  foot  of  a  tower  surmounted  by 
a  pole,  the  angle  of  elevation  of  the  top  of  the  tower  is  35°  and  that  of 
the  top  of  the  pole  is  47°.     Find  the  length  of  the  pole. 


DIVISION   OF  A  SEGMENT  179 

C,     Proportional  Division  of  a  Segment 
Proposition  XIX.     Theorem 

304.   Parallel  lines   intercept   proportional  segments   on    all 
transversals. 

X 


Hypothesis.     Parallels  AB,  CD,  EF,  and  GH  intercept  seg- 
ments a,  b,  and  c,  on  XY  and  d,  e,  and/  on  ZW,  respectively. 

Conclusion.  -  =  -  =  -. 

d     e     f 

Suggestions.  —  1.    Draw  TK II  Z  W.    Let  the  lU  intercept  the  segments 
r,  s,  and  t  on  TK. 

2.  Compare  r,  s,  and  t,  with  d,  e,  and  /,  respectively. 

3.  In  A  TKL,  compare  a  :  r  with  TL  :  TK. 

4.  In  A  TKL,  compare  c  :  t  with  TL  :  TK 

5.  In  A  TMX,  compare  a  :  r  with  b  :  s. 

6.  Then  compare  a  :r,  c  :t,  and  b  :  s. 

Complete  tlie  proof,  using  the  facts  obtained  in  step  2. 

Ex.  121.    Divide  a  segment  into  parts  proportional  to  any  number  of 
given  segments. 


Given  segment  AB,  and  segments  w,  n,  and  p. 
Required  to  divide  AB  into  segments  x,  y,  and  z,  so  that 

—  =  ^  =  -. 
m     n     p 

Suggestion. — Base  the  construction  on  Proposition  XIX. 


180  PLANE    GEOMETRY  —  BOOK   III 

Ex.  122.  Divide  a  segment  6  in.  in  length  into  segments  proportional 
to  2,  3,  and  4. 

Ex.  123.  Construct  a  triangle  similar  to  a  given  triangle,  having 
given  its  perimeter. 

Ex.  124.  A  line  parallel  to  the  bases  of  a  trapezoid,  passing  through 
the  intersection  of  the  diagonals,  and  terminating  in  the  non-parallel 
sides,  is  bisected  by  the  diagonals. 

Note.  — Supplementary  Exercises  45  to  46,  p.  293,  can  be  studied  now. 

D.     External  and  Harmonic  Division  of  a 
Segment 

305.  External  Division  of  a  Segment.  If  P  is  a  point  on 
line  AB  but  not  located  between  A  and  B,  then  P  divides  AB 
externally  into  segments  AP  and  PB. 

FA  B      A^  B         P 


Fig.  1 

The  following  justification  is  given  for  calling  AP  and  PB  segments 
of  AB.  Direction  on  a  segment  may  be  indicated  by  reading  the  seg- 
ment from  its  beginning  point  to  its  end  point ;  thus 

AB  =A >-  B,  and  BA  =  A  -^ B. 

Direction  from  left  to  right  is  regarded  as  positive  and  from  right  to 
left  as  negative.     Hence  BA  =—  AB  and  AB  +  BA  =  0. 

Clearly  then  the  algebraic  sum  of  JP  and  PB  in  Fig.  1  equals  AB, 
for      AP+  PB=AP+  PA-\-  AB      and      AP  +  PA  ^  0. 

The  consequence  is  that  the  algebraic  sum  of  AP  and  PB  is  AB,  no 
matter  where  P  is  located  on  the  line  AB.     (See  figures  below.) 

A.  P        B    p  A  B     A  B      P 


Fig.  2 
Note.  —  Prove  that  AP  +  PB  =  AB  in  each  case  in  Fig.  2. 

;^  BCD  E^-  125.     What  segment  represents  the  alge- 

'  •"•  braic  sum  of :     (a)  AB  +  J5C?     (b)  AD -{■DC'} 

(c)  BA^-  AG  -\-  CB'} 


DIVISION  OF  A  SEGMENT 
Proposition  XX.     Theorem 


181 


306.  In  any  triangle,  the  bisector  of  an  exterior  angle  at  any 
vertex  divides  the  opposite  side  externally  into  two  segments  whose 
ratio  equals  the  ratio  of  the  two  adjacent  sides  of  the  triangle. 


D  B 

Hypothesis.   AD  bisects  ext.  Z  BAE  of  A  ABC,  meeting  CB 

extended  at  D. 
Conclusion.  BD:DC=  BA :  AC. 

Proof.    1.  Draw  BF  II  DA,  meeting  AC  at  F. 
2.   Then  BD  :  DC  =  FA:  AG. 

It  remains  to  prove  that  FA  =  BA.  Try  to  prove  that  Z3=Z4, 
using  the  hypothesis  and  construction.     Proof  left  to  the  pupil. 

Note.  —  The  converse  of  Proposition  XX  is  also  true.  It  may  be 
proved  by  laying  off  AF  =  AB. 

Ex.  126.  The  sides  of  a  triangle  are  AB  =  5,  BC  =  7,  and  CA  =  8  ; 
find  the  segments  into  which  side  8  is  divided  by  the  bisector  of  the  ex- 
terior angle  at  the  opposite  vertex. 

307.  A  segment  is  divided  harmonically  if  it  is  divided  inter- 
nally and  externally  into  seg-    j_  X      B  Y 
ments  having  the  same  ratio.                  i            \         I 

Thus. MAB=\,AX=  -,  and  JIF  =  %  then  ^  =  ^,  since  each 
4  2  JlB      YB 

ratio  equals  -.     Hence  Xand  F  divide  AB  harmonically. 

Ex.  127.  Prove  that  the  bisector  of  an  interior  angle  of  a  triangle 
and  the  bisector  of  the  exterior  angle  at  the  same  vertex  divide  the  op- 
posite side  harmonically. 

Ex.  128.  If  Xand  F divide  AB  harmonically,  then  A  and  B  divide 
Xr  harmonically.    (Use  §  263.    Verify  afterwards  for  the  figure  in  §  307. ) 


182 


PLANE    GEOMETRY  — BOOK   III 


E.    Numerical  Relations  among  Segments  of  a 

Triangle  ./> 

308.  The  Projection  of  a  Point  upon  a 
given  line  is  tlie  foot  of  the  perpendicu- 
lar drawn  from  the  point  to  the  line.  j^.  ^^^ 

Thus,  B  is  the  projection  of  P  on  AB. 


309.   The    Projection    of    a   Segment 

upon  a  given  line  is  the  distance  between 
the  projections  of  its  end-points. 

Thus,  the  projection  of  AB  on  line  CD  is 
A'B'. 

The  symbol  "  jo^^ "  is  read  "the  projec- 
tion of  AB  on  CDr 


C  A 


B   D 


Ex.  129.  Draw  a  segment  AB  and  also  four  straight  lines  not  paral- 
lel to  AB  but  also  not  crossing  AB. 

(a)  Determine  the  projection  of  AB  on  each  of  the  straight  lines. 

(b)  Are  the  projections  all  of  the  same  length  ? 

Ex.  130.  Draw  an  acute  scalene  triangle.  Show  by  means  of  a 
drawing  the  projection  of  the  shortest  side  upon  each  of  the  other  sides. 

Ex.  131.     Repeat  the  preceding  exercise  for  an  obtuse  triangle. 

Ex.  132.  Draw  an  obtuse  triangle.  Obtain  the  projection  of  the 
longest  side  upon  each  of  the  other  sides. 

Ex.  133.  If  AB,  extended,  makes  an  acute  angle  with  a  line  m, 
prove  that  p^^  is  less  than  AB. 

Ex.  134.    If  AB  II  ?7i,  how  does  p^  compare  with  segment  AB  ? 

Ex.  135.     If  AB  ±  m,  what  is  the  length  oip'l^  ? 

Ex.  136.  What  part  of  the  base  of  an  isosceles  triangle  is  the  pro- 
jection upon  the  base  of  one  of  the  equal  sides  ? 

Ex.  137.  If  the  equal  sides  of  an  isosceles  trapezoid  be  projected 
upon  the  lower  base,  the  projections  are  equal. 

Ex.  138.  Draw  a  right  triangle  and  draw  the  median  to  the  hypotenuse. 
Prove  that  the  projection  of  the  median  upon  either  leg  of  the  triangle  is 
one  half  of  that  leg. 


NUMERICAL  RELATIONS  AMONG  SEGMENTS      183 


Propositiox  XXI.     Theorem 

310.  In  any  triangle,  the  square  of  the  side  opposite  an  acute 
angle  is  equal  to  the  sum  of  the  squares  of  the  other  two  sides, 
minus  twice  the  product  of  one  of  these  sides  and  the  projection  of 
the  other  upon  it. 


Fig.  1 

Hypothesis.     In  A  ABC,  Z  5  is  acute. 
Conclusion.  6^  =  a^  +  c'^  —  2  a  •  ^^. 

Proof.     1.   Draw^Z)±50.     Then  ^i>=i9^=i). 

2.  In  Fig.  1,  ?>2  =  /i2  4-  DC".  Why  ? 

3.  But  i)0  =  a  -  p,  and  /i^  =  c^  -  p\  Why  ? 

4.  .-.  62  =  c2  -  />2  -f  (a  -  j))\  Ax.  2,  §  51 

(Complete  the  proof.) 

Note  1.  —  A  similar  proof  may  be  given  from  Fig.  2.  In  Fig.  2,  DC  = 
p  —  a. 

Note  2.  —  The  conclusion  of  Proposition  XXI  is  a  formula  connecting 
the  three  sides  of  a  triangle  with  the  projection  of  one  side  upon  one  of  the 
otlier  two  sides.  Altogether  four  different  numbers  are  involved.  Hence, 
when  three  of  these  numbers  are  known,  the  fourth  may  be  determined  by 
substituting  in  the  formula  and  solving  the  resulting  equation.  In  the 
right  member,  there  appear  the  squares  of  two  sides  and  the  projection  of 
one  of  these  upon  the  other ;  in  the  left  member,  there  appears  the  square 
of  the  third  side. 

Ez.  139.     Determine  : 

(a)  ;)«  when  a  =  13,  &  =  14,  and  c  =  15. 
(6)  6   when  a  =  10,  c  =  12,  and  jo«  =  9. 

(c)  c    when  a  =  11,  6  =  16,  and  p^  =  7. 

(d)  a   when  6  =  18,  c  =  12,  andp^  =  4. 


184  PLANE    GEOMETRY  —  BOOK   III 

Proposition  XXII.     Theorem 

311.  In  any  triangle  having  an  obtuse  angle,  the  square  of  the 
side  opposite  the  obtuse  angle  is  equal  to  the  sum  of  the  squares  of 
the  other  two  sides,  plus  twice  the  product  of  one  of  these  sides  and 
the  projection  of  the  other  side  upon  it. 

A 


Hypothesis.     In  A  ABC,  Z  (7  is  an  obtuse  Z. 
Conclusion.  c^=^a''+b^-\-2  a-pl 

Proof.     1.   Draw  AD  ±  BC  extended.     Then  p^c  =CD=p. 

2.  Then  in  A  ABD,  ^  =  li^  +  Bff-  Why  ? 

3.  But  7i2  =  ?>2  _p2 .  and  BD  =  a+p. 

4.  .-.  c'  =  (b''-p'')^{a+pf. 
(Complete  the  proof.) 

312.     Cor.     If  a,  b,  and  c  are  the  sides  of  a  triangle : 
Z  ^  is  acute         if  a^  <  b'^^c}\ 
Z.Ais  obtuse       if  a^  >  6^  +  c^ ; 
Z  A  is  Si  right  Z  if  a^  =  6^  +  c\ 
(Proof  is  indirect  in  each  case.) 

Ex.  140.  Is  the  greatest  angle  of  the  triangle  whose  sides  are  8,  9, 
and  12  acute,  right,  or  obtuse  ? 

Ex.  141.  Is  the  greatest  angle  of  the  triangle  whose  sides  are  12,  35, 
and  37  acute,  right,  or  obtuse  ? 

Ex.  142.  Prove  that  the  sum  of  the  squares  of  the  diagonals  of  a 
parallelogram  equals  the  sum  of  the  squares  of  the  sides  of  the  parallelo- 
gram.    (Use  §  310  and  §  311.) 

Ex.  143.     If  AB  and  A  C  are  the  equal  sides  of  an  isosceles  triangle, 
and  BD  is  drawn  perpendicular  to  AC,  prove 
2AC  X  CD  =  BC^. 
Note.  — Supplementary  Exercises  47  to  51,  p.  293,  can  be  studied  now. 


NUMERICAL  RELATIONS  AMONG  SEGMENTS        185 


313.   When  the   three  sides  of  a  triangle  are   known,  the 
altitude  to  each  side  can  be  computed. 

A 


1.   Assume  AD  =  /i„,  and  Z  ^  to  be  an  acute  angle. 


2. 
3. 
4. 
5. 

6. 

7. 
8. 


62=;a2  +  c2-2a  .p^  or  ¥  =  a^ -\' c^ - 2 ap. 


.-.  p  = 


2a 


/l2  =  c2-p2=(c4-p)(C-p). 


K= 


;+        2a       r ^^~"J 

2  a  JL  2  a  J 


4a2 


^  (g  -h  c  4-  &) (q  +  c  -  6)(&  4-  g  -  c) (&  -  g  H-  c) 

4g2 

9.   Let  g4-6  +  c  =  2s. 

10.  .-.  g  +  6-c  =  2s-2c  =  2(s-c). 

Similarly,     6-|-c  —  g  =  2(s  —  g);  and  c -f- g  —  6  =  2 (s  —  6). 

-,.      .  ^2  _  2/?.  2  (8 -6)  '2.  (a-c)  '2{s-a) 
J. J..   . .  /i^  — 

4g* 

_  4  8(8  —  g)(/t  —  6)(8  —  c) 


12.    .-.  ^„  =  -Vs(8-g)(s-6)(s-c) 
g 


Similarly, 
and 


K  =  -y/s{s-a)  (s  -  6)  (s  -  c)  ; 

0 

o        

h^  =  -  Vs(«  —  g)  (s  —  6)  (s  —  c)- 
c 


Ex.  144.     Find  the  three  altitudes  of  the  triangle  whose  sides  are  13, 
14,  and  16,  getting  the  results  correct  to  one  decimal  place. 


186  PLANE    GEOMETRY  —  BOOK   III 

Proposition  XXIII.     Theorem 

314.  In  any  triangle,  the  sum  of  the  squares  of  two  sides 
equals  twice  the  square  of  half  the  third  side  plus  twice  the  square 
of  the  median  drawn  to  that  side. 

O 


L .^..A _J 

Hypothesis.     In  A  ABC,  CD  is  the  median  to  side  AB. 
Conclusion.  a^-{-¥  =  2  ^^  Y  +  2  mf. 

Note.  — A  ABC  is  either  a  rt.  Z,  an  acute  Z,  or  an  obtuse  Z.  When 
it  is  a  rt.  Z,  the  proof  is  quite  easy. 

Proof.  1.  Assume  that  Z  ABC  is  obtuse  and  hence  Z  BBC 
is  acute. 

2.   Draw  CE  ±  AB,  so  that  BE=pll. 

(Complete  the  proof.) 

Suggestion.  —  Determine  a^  from  ABCDhj^  310 ;  62  from  AACDhy^  311 ; 
then  add,  so  as  to  obtain  a^-{-  b^. 

Note.  —  By  Proposition  XXIII,  it  is  possible  to  determine  the  three 
medians  of  a  triangle  when  the  three  sides  of  the  triangle  are  known. 

315.  Cor.  The  difference  between  the  squares  of  tzvo  sides  of 
a  triangle  equals  tivice  the  product  of  the  third  side  and  the  pro- 
jection of  the  media}!  upon  that  side. 

Con.  b^-a^  =  2G'  p'^c. 

Suggestion.  —  Determine  6^  and  a^  and  then  subtract  the  value  of  a^  from 
that  of  62. 

Ex.  145.  Determine  m^  when  &  =  12,  c  =  16,  and  a  =  20.  Deter- 
mine also  m^  and  rric. 

Ex.  146.  Prom  the  conclusion  of  §  314  derive  a  formula  for  m^  in 
terms  of  a,  6,  and  c. 


NUMERICAL  RELATIONS  AMONG  SEGMENTS      187 


Proposition  XXIV.     Theorem 

316.  In  any  triangle,  the  product  of  two  sides  equals  the  product 
of  the  diameter  of  the  circumscribed  circle  and  the  altitude  upon 
the  third  side. 


Hjrpothesis.     A  ABC  is  inscribed  i  n  O  0 ;  AD  is  a  diameter 
of  O  0;  AE  =  h,. 

Conclusion.  b  -  c  =  d  -h^. 

Analysis  and  proof  left  to  the  pupil.    See  analysis  of  §  283. 

2 


317.   By  §  313,        h,  =  -  Vs(s  -  a)(s  -  b)(s  -  c). 


2d 


Hence,  by  §  316,      b  -  c  =  —  Vs(s  —  a){s  —  b){s  —  c). 

a 


.'.  2  dVs(s  —  a){s  —  b)(s—c)  =  abCj 
d=  ^^^ 


2Vs(s-a)(s-b){s-c) 
Hence,  when  the  sides  of  a  triangle  are  known,  the  diameter  of 
the  circumscribed  circle  can  be  computed. 

Ex.  147.  Determine  the  diameter  of  the  circle  circumscribed  about 
the  triangle  whose  sides  are  13,  14,  and  15. 

Ex.  148.  If  two  adjacent  sides  and  one  of  the  diagonals  of  a  parallelo- 
gram are  7,  9,  and  8  respectively,  find  the  other  diagonal.  (§  314.) 

Ex.  149.  The  sides  AB  and  ^C  of  A  ABC  are  16  and  9  respectively, 
and  the  length  of  the  median  drawn  from  C  is  11.    Find  side  BC.  (§  314.) 

Ex.  150.  If  the  sides  oi  £\ABC  are  10,  14,  and  16,  find  the  lengths  of 
the  three  medians.  Determine  also  the  diameter  of  the  circumscribed 
circle. 


188 


PLANE    GEOMETRY  —  BOOK   III 


Proposition  XXV.     Theorem 

318.  In  any  triangle,  the  product  of  any  two  sides  is  equal  to 
the  product  of  the  segments  of  the  third  side  formed  by  the  bisector 
of  the  opposite  angle,  plus  the  square  of  the  bisector. 


■~-v^ 


Hypothesis.     AD  bisects  Z  J.  of  A  ABC,  meeting  BC  at  D. 
(Let  BD=r,  and  DC  =  p.) 

Conclusion.  b-c^tl  +  r-p. 

Proof.     1.    Circumscribe  a  circle  about  A  ABC. 

Extend  AD  to  meet  the  circle  at  E.     Draw  CE. 


2. 

A  ABD  ~  A  AGE. 

Give  full  proof. 

3. 

•••;^=i'°^'"'= 

AE '  t^. 

Give  full  proof. 

4. 
5. 
6. 

7. 

•••  bc=^t\'(tj,  +  s). 
.:bc  =  t\  +  tj,'S. 
But  t^-  s  =  r  'p. 

.'.  bc  =  t\-\-r'p. 

{AE  =  t^  +  s) 
§283 

Note.  —  This  proposition  makes  it  possible  to  compute  tlie  bisectors  of 
the  three  angles  of  a  triangle  when  the  three  sides  of  the  triangle  are 
known. 

Ex.  151.     If  c  =:  4,  6  =  5,  and  a  =  6,  find  Ia- 

Suggestions.  —  1.  It  is  necessary  to  find  r  and  p  first.  This  may  be  done 
by  using  §  270. 


r     ^4 
6-r     5 
2.   Then  substitute  in  the  conclusion  of  §  318 


Whence  r=?  and  p  =  6  —  r  =  ? 
mclusion  of  §  318. 
Note.  —  Supplementary  Exercises  52  to  56,  p.  294,  can  be  studied  now. 


MISCELLANEOUS  EXERCISES  189 

Miscellaneous  Exercises 
Ex.  152.     The  vertices  of  quadrilateral  ABCD  are  joined  to  a  point 
0  lying  outside  the  quadrilateral.     Points  A',  B' ,  C,  and  D'  are  taken 
on  OA,  OB,  OC,  and  OD,  respectively,  so  that  A'B'  II  AB,  B'C  II  BC\ 
and  CD'  II  CD.    Prove  A'D'  II  AD. 

Ex.  153.  Two  circles  are  tangent  externally  at  point  C.  Through 
C,  a  straight  line  is  drawn,  meeting  the  first  circle  at  A  and  the  second  at 
D ;  another  straight  line  through  C  meets  the  first  circle  at  B  and  the 
second  at  E.    Prove  AC:  CD  =  BC:  CE. 

Suggestion. — Draw  the  common  tangent  at  C,  and  also  chords  AB  and 
ED. 

Ex.  154.  If  P  and  8  are  two  points  on  the  same  side  of  line  OX 
such  that  the  perpendiculars  PB  and  /ST  drawn  to  OX  have  the  same 
ratio  as  OB  and  OT,  then  points  0,  P,  and  SWq  in  a  straight  line. 

Suggestion.  — Frove  L  ROP  =  L  TOS  by  proving  A  OPR  ~  A  OST. 

Ex.  155.  If  two  parallels  are  cut  by  three  or 
more  straight  lines  passing  through  a  common 
point,  the  corresponding  segments  are  propor- 
tional. 

Prove  ^-=.^  =  ^^.  -y 

A'B'      B'C      CD'  /' 

Ex.  156.  If  three  transversals  intercept  proportional  lengths  on  two 
parallels,  the  transversals  meet  at  a  point. 

Suggestion. — Let  A' A  and  B'B  meet  at  0  and  draw  OC  and  OC";  then 
prove  A  OBC&nd  OB'C  similar.     (Fig.  Ex.  155.) 

Ex.  157.  Derive  a  formula  for  the  altitude  to  the  base  of  an  isosceles 
triangle  if  the  base  is  b  and  the  equal  sides  are  each  a.  By  means  of  the 
formula  determine  the  altitude  when  :  (^a)  a  =  12  and  b  =6;  (b)  a  = 
15  and  6=7. 

Ex.  158.  Find  the  length  of  the  common  external  tangent  to  two 
circles  whose  radii  are  11  and  18,  if  the  distance  between  their  centers 
is  25. 

Suggestion.  —  See  the  figure  for  Problem  2,  §  236. 

Ex.  159.  If  BE  and  CF  are  the  medians  drawn  from  the  extremities 
of  the  hypotenuse  of  right  triangle  ABC,  prove  4  BE^  +  4  CY'^  =  5  BC'\ 

Ex.  160.  Prove  that  the  projections  of  two  parallel  sides  of  a  paral- . 
lelogram  upon  either  of  the  other  sides  are  equal. 


190  PLANE    GEOMETRY  —  BOOK   III 

Ex.  161.  BC  is  the  base  of  an  isosceles  triangle  ABC  inscribed  in 
a  circle.  If  a  chord  AD  is  drawn,  cutting  BG  at  E,  prove  AB^  =  AE^  + 
BE  X  CE. 

Suggestions.  —  1.  The  proof  is  like  that  for  §  318. 
2.   Prove  A  ABD  ~  A  ABE. 

Ex.  162.  Prove  that  the  non-parallel  sides  of  a  trapezoid  and  the  line 
joining  the  middle  points  of  the  parallel  sides,  if  extended,  meet  in  a  com- 
mon point. 

Review  Questions 

1.    What  is  meant  by  taking  a  proportion  by   (a)  inversion?    (b) 
composition  ?  (c)  alternation  ? 

.    2.   Complete  the  following  theorem  :  "If  the  product  of  two  numbers 
equals  the  product  of  two  other  numbers,  one  pair,  .-  " 

3.  Define:    (a)    mean  proportional;    (6)    fourth   proportional;   (c) 
similar  polygons ;  (d)  ratio  of  similitude. 

4.  Are  mutually  equiangular  triangles  similar  ? 
Are  mutually  equiangular  polygons  similar  ? 

5.  State  all  of  the  theorems,  by  which  two  triangles  can  be  proved 
similar. 

6.  How  do  you  select  the  homologous  sides   of  similar  triangles  ? 
What  do  you  know  about  them  ? 

7.  What  do  you  know  about  the  ratio  of  homologous  altitudes  of 
similar  triangles  ? 

What  about  the  ratio  of  the  perimeters  of  similar  triangles  ? 
What  about  the  ratio  of  the  areas  of  similar  triangles  ? 

8.  What  is  the  Pythagorean  theorem  ? 

9.  Find  the  mean  proportional  between  5  and  15. 
Construct  the  mean  proportional  between  segments  r  and  s. 

10.  What  are  the  two  conclusions  which  follow  from  the  hypothesis 
that  the  altitude  is  drawn  to  the  hypotenuse  of  a  right  triangle  ? 


BOOK  IV 


AREAS   OF  POLYGONS 


319.  A  polygon,  being  a  closed  line  (§  7),  incloses  a  limited 
portion  of  the  plane. 

In  measurement  theorems,  the  words  "rectangle,"  "parallelogram," 
*' polygon,"  etc.,  mean  the  surface  within  the  figure  mentioned. 

320.  The  Area  of  the  surface  within  a  closed  line  is  the 
ratio  of  the  surface  to  the  unit  of  surface  measure. 

Thus,  in  the  adjoining  figure,  if  the  unit  of  surface 
is  one  small  square,  the  area  of  the  rectangle  is  30. 

It  has  become  customary,  when  speaking  of  the 
area  of  a  figure,  to  mention  at  once  the  unit  of  sur- 
face ;  thus,  in  the  foregoing  example,  it  is  customary 
to  say  that  the  area  is  30  small  scjuares.  Remember, 
however,  that  the  area  is  30.  -^ 

321.  The  usual  Unit  of  Surface  is  a  square  whose  side  is 
some  linear  unit :  as,  a  square  inch  or  a  square  centimeter. 
In  this  text,  it  will  be  assumed  that  the  unit  always  is  such  a 
square  unit. 

Ex.  1.  In  the  following  figures,  assume  that  the  unit  of  surface  is  a 
small  square,     (a)  What  is  the  exact  area  of  Figs.  1  and  2  ? 


i._.H.._4 — J.-_4--- 

1...4.— ^--4._.^._- 
— -i— T— I-— f-- -f— 


Fig.  1 


FiQ.  2 


191 


192 


PLANE    GEOMETRY  —  BOOK   IV 


2               7  -     /^-.                  ^"     ""—^ 

L                   t       /         ^^                t               A 

i                7     V             ^^        V               -^ 

'        Jl     f      ^                ^^Jli   -^        ^ 

Fig.  3 


Fig.  4 


Fig.  5 


(?))  What  is  the  approximate  area  of  Figs.  3,  4,  and  5  ? 
(Include  the  square  in  the  area  if  half  or  more  than  half  of  it  lies 
within  the  figure  ;  do  not  include  it  otherwise.) 

322.  Two  limited  portions  of  a  plane  are  Equal  (=)  if  their 
areas  are  equal  when  they  are  measured  by  the  same  unit. 

Since  the  test  of  the  equality  of  two  figures  is  the  equality  of  two 
numbers,  the  usual  axioms  apply  when  equal  figures  are  added  or  sub- 
tracted, or  when  they  are  multiplied  or  divided  by  the  same  number. 

Thus,  if  equal  figures  are  added  to  equal  figures,  the  sums  are  equal  ; 
also  halves  of  equal  figures  are  equal. 

Ex.  2.     Of  the  figures  in  Ex.  1,  are  any  two  or  more  equal  ? 

323.  Two  congruent  figures  are  necessarily  equal,  hut  two 
equal  figures  are  not  necessarily  congruent. 

Also,  two  figures  which  consist  of  parts  which  are  respectively  con- 
gruent are  equal. 

Thus,  the  parallelogram  and  the  kite- 
shaped  figure  made  from  it  by  placing       mm^^^ 
the  two  triangles  together  as  hi  the  fig-    //////////§i 
ure  adjoining  are  equal. 

Ex.  3.     If  E  is  the  mid-point  of  one  of  the  non- 
parallel  sides  of  trapezoid  ABCD,  and  a  parallel  to 
AB  drawn  through  E  meets  BC  extended  at  F  and 
AD  at  G,  prove  that  parallelogram  ABFG  is  equal    ^ 
to  trapezoid  ABGD. 

Suggestion.  — Fvoye  A  CEF  ^  A  GED,  and  apply  §  323. 

Ex.  4.     In  the  adjoining  figure,  D  and  E  are 
the  mid-points  of  ^B  and  ^C;   AJA.BG;   BF 
±  DE  extended  at  F;  CG  ±  DE  extended  at  G.         ^.  ,^ 
Prove  th at  A  ^^  O  equals  un  BFGC. 

Suggestion.  —  Prove     A  BDF  ^  A  DAH,     and 
A  CEG  ^/\AEH. 


\L 


AREAS  OF  POLYGONS 


193 


Ex.  5.  In  the  adjoining  figure,  E  and  F  are 
the  mid-points  of  sides  AB  and  CD  of  trapezoid 
ABCD  ;  XY  and  Z  \V  are  drawn  througli  E  and 
F  respectively  ±  AD,  meeting  BC  extended  at  X 
and  irrespectively.    VvovQih2Lt  XYWZ=ABCD.     "^  .^     " 

Ex.  6.  Let  K  be  the  mid-point  of  side  5Cand  if  the  mid-point  of 
side  AD  of  CD  ABCD  ;  let  FE,  drawn  through  the  mid-point  G  of  KH, 
intersect  BC  and  AD  at  F  and  E  respectively.  Prove  that  FE  divides 
ABCD  into  two  equal  quadrilaterals. 

Note.  —  Supplementary  Exercises  1-3,  p.  294,  can  be  studied  now. 


MEASUREMENT  OF  RECTANGLES 

324.  The  Dimensions  of  a  rectangle  are  the  Base  and 
Altitude. 

325.  Area  of  a  Rectangle.  (Informal  treatment.)  If  the 
base  of  a  rectangle  measures  6  and  its  altitude  5  linear  units, 
the  area  is  evidently  6  x  5  or  30  surface  units. 

If  the  base  measures  6  units  and  the  alti- 
tude measures  3^  units,  the  area  is  evidently 
6  X  3.5  or  21  surface  units. 

These  two  examples  suggest  the  theorem  : 

The  number  of  surface  units  in  the  area  of  a  rectangle  is  the 
product  of  the  number  of  linear  units  in  its  base  and  the  number 
in  its  altitude.  More  'briefly,  this  theorem  is  expressed :  the 
area  of  a  rectangle  is  the  product  of  its  base  and  its  altitude. 

The  theorem  is  proved  in  the  following  three  propositions. 


T     w 


— I— r----|-T 
J_.L4.J__I 

.4-1-^-1-4 


X 


326.  Comparison  of  Rectangles.  Rectangles  may  be  com- 
pared without  computing  their  areas. 

Two  rectangles  having  equal  bases  and  altitudes  are  equal,  for 
it  is  evident  that  they  can  be  made  to  coincide  by  superposition. 


194 


PLANE    GEOMETRY  —  BOOK   IV 


Pkoposition  I.     Theokem 

327.   Two   rectangles  having   equal  altitudes   are  to 
each  other  as  their  bases. 


K 


c 


D 


E 


G 


H 


Hypothesis.      Rectangles   ABCD   and   EFGH  have   equal 
altitudes  AB  and  EF,  and  bases  AD  and  EH,  respectively. 

Conclusion.  ABCD  ^  AD  ^ 

EFGH     EH 

Case  I.     Assume  that  AD  and  EH  are  commensurable. 

§211 
Proof.     1.   Let  AK,  a  common  measure  of  AD  and  EH,  be 
contained  in  AD  5  times  and  in  EH  3  times.     Draw  Js  to  AD 
and  EH  at  the  points  of  division. 

2.    Then   ABCD   and  EFGH  are  divided  into  equal   rec- 
tangles. Why  ? 
(Complete  the  proof.) 

Suggestions.  —  What  is  the  value  of  ^-^  ?  of  -  ?    Then  compare  these 

ratios.  ^^         ^^^H 

Case  IL     When   AD   and  EH  are   incommensurable,  the 
theorem  is  still  true.     The  proof  is  given  in  §  425. 

328.   Cor.     Two   rectangles  having   equal   bases  are   to  each 
other  as  their  altitudes. 

Ex.  7.     Construct  a  rectangle  which  will  be  three  times  a  given  rec- 
tangle ; '  also  one  which  will  be  three  fourths  a  given  rectangle. 

Ex.  8.     Two  rectangles  M  and   T  have  equal  bases  b  and  altitudes 
r  and  s  respectively.     What  is  the  ratio  of  J^f  to  2'? 

Note.  —  Supplementary  Exercises  4  to  5,  p.  295,  can  be  studied  now. 


AREAS   OF  POLYGONS  195 

Proposition  II.     Theorem 

329.   Two  rectangles  are  to  each  other  as  the  products 
of  their  bases  by  their  altitudes. 


M 


B 


h' 

Hypothesis.     Rectangle  M  has  base  h  and  altitude  a ;  rec- 
tangle N  has  base  V  and  altitude  a'. 

Conclusion .  —  =  -—  • 

N     a'b' 

Proof.     1.   Let  rectangle  R  have  base  6'  and  altitude  a. 

2.  .-.  —  =  — ,  since  M  and  E  have  equal  altitudes.     Why? 

7?       n 

3.  Also      —  =  — ,  since  R  and  iVhave  equal  bases.      Why? 

,  M  ^R      ab    ^    M      ah  .^ttj.     ^ 

4.  •••  —  X -T^=— :t-.,  or  —  =  -^rT-  Why? 


.    =-x-  =  —    or^=-^ 
"  R      N     a'b''       N      a'b' 


Ex.  9.     What  is  the  ratio  of  rectangles  B  and  S  if  their  dimensions 
are  as  follows  ? 


(A 

) 

(B) 

(C) 

Ji 

s 

li 

s 

It 

*s- 

Altitude 

k 

X 

Altitude 

4 

6 

Altitude 

12 
20 

15 
18 

Base 

w 

y 

Base 

10 

16 

Base 

Ex.  10.  If  i?,  8,  T,  and  X  are  rec- 
tangles having  the  dimensions  indicated  in 
the  adjoining  table,  determine  the  ratio  of 
each  rectangle  to  each  of  the  others. 

(Thus,  determine  B  :  8,  B  :  T,  Qiz.) 


Altitl-de 

Base 

B 

10 

6 

8 

5 

12 

T 

10 

8 

X 

10 

12 

196 


PLANE    GEOMETRY  —  BOOK   IV 


Proposition  III.     Theorem 

330.   The  area  of  a  rectangle  is  the  product  of  its 
base  and  altitude. 


N 


Hypothesis.     Eectangle  M  has  altitude  a  and  base  b. 

Conclusion.  Area  oi  M=ab. 

Proof.     1.    Let  square  ^be  the  unit  of  surface  measure. 

2.  Then  area  of  M=  the  ratio  of  Mto  N.  §  320 

M_     ab 

N~lxl' 
4.  .-.  area  of  M=  ab. 


3. 


Why? 


Note.  —  Remember  that  this  theorem  means  that  the  number  of 
square  units  in  the  area  equals  the  product  of  the  number  of  linear  units 
in  the  base  by  the  number  in  the  altitude.  A  similar  interpretation 
must  be  given  for  each  of  the  measurement  theorems  of  this  Book. 

Ex.  11.  A  business  corner  50  ft.  x  120  ft.  is  valued  at  .$9000.  What 
is  the  value  per  square  foot  ? 

Ex.  12.     The  area  of  a  square  is  590.49  sq.  ft.     Find  its  perimeter. 

Ex.  13.  A  rectangle  has  the  dimensions  30  ft.  and  120  ft.  Compare 
its  perimeter  with  that  of  an  equal  square. 

Ex.  14.  An  ordinary  eight-room  house  costs  approximately  $4.75 
per  square  foot  of  ground  covered  by  it.  What  is  the  approximate  cost 
of  a  house  27  ft.  x  36  ft.? 

Ex.  15.  The  area  of  a  rectangle  is  147  sq.  ft.  Its  base  is  three  times 
its  altitude.     What  are  its  dimensions  ? 

Ex.  16.  What  are  the  dimensions  of  a  rectangle  whose  area  is  168  sq. 
ft.  and  whose  perimeter  is  58  ft  ? 

Suggestion.  —  Let  the  base  =  x  and  the  altitude  =  y.  Form  two  equations 
and  complete  the  solution  algebraically. 

Ex.  17.  What  is  the  length  of  the  diagonal  of  a  rectangle  whose  area 
is  2640  sq.  ft.,  if  its  altitude  is  48  ft.? 


AREAS   OF   POLYGONS 


197 


Proposition  IV.     Theorem 

331.   TJie  area  of  a  parallelogram  equals  the  product 
of  its  base  and  altitude. 

E        B      


Hypothesis.     ABCD  is  a  parallelogram. 

Its  altitude  DF=  a :  its  base  AD 


b. 


Conclusion.     Area  of  ABCD  =  ah. 

Proof.     1.   Draw  AE  \\  DF,  meeting  BC  extended  at  E. 

2.  .-.  AEFD  is  a  rectangle.  Why  ? 

3.  A  AEB  =  A  FCD.         Give  the  full  proof. 

4.  .-.  O  ABCD  =  □  AEFD.  Why  ? 

5.  But  area  of  □  AEFD  =  ah.  Why  ? 

6.  .-.  area  of  ABCD  =  ah.  Ax.  1,  §  51 

332.   Corollaries.     Let  O  P^  have  base  ftj  and  altitude  a/; 
and  O  P.,  have  base  &2  and  altitude  ag. 

(1)  Parallelograms  having  equal  hases  and  equal  altitudes  are 
equal. 

(2)  Two  parallelograms  are  to  each  other  as  the  products  of 
their  hases  hy  their  altitudes. 

For,  since  O  Pi  =  ai6i  and  O  r^  =  a-rbi,  then  ^^  =  ^  • 

O  P2      0262 

(3)  Parallelograms  having  equal  altitudes  are  to  each  other  as 
their  hases. 

For,  in  (2),  if  ai  =  ao,  then  O  Pi :  O  P2  =  61 :  62. 

(4)  Parallelograms  having  equal  bases  are  to  eojch  other  as 
their  altitudes. 


198 


PLANE    GEOMETRY  —  BOOK   lY 


Ex.  18.     What  is  the  area  of  OB,  of  O  S,  and  ofUT? 

(a)  O  B  has  altitude  4  in.  and  base  9  in. 

(b)  JU  S  has  altitude  15  ft.  and  base  20  ft. 

(c)  CJ  T  has  altitude  3  a;  in.  and  base  11  y  in. 

Ex.  19.  What  is  the  altitude  of  a  parallelogram  whose  area  is  56 
sq.  in.,  if  its  base  is  14  in.  ? 

Ex.  20.  Construct  a  parallelogram  equal  to  twice  a  given  parallelo- 
gram. 

Ex.  21.  Construct  a  rectangle  equal  to  two  thirds  a  given  parallelo- 
gram. 

Ex.  22.  Divide  a  parallelogram  into  two  equal  parallelograms ;  into 
four  equal  parallelograms. 

Ex.  23.  What  is  the  ratio  oi  O  P  to  O  B  \i  the  base  of  each  is  10 
in.  and  the  altitudes  are  5  in.  and  8  in.  respectively  ? 

Ex.  24.  Construct  a  CJ  ABCD  having  AB  =  3  in.  and  BG  =  4  in., 
and  having  :  (a)  ZB  =  30^;  (6)  ZB  =  45°.  (c)  Determine  the  area  of 
each  of  the  parallelograms. 

Ex.  25.     The  base  of  A  ABC  is  10  and  the  alti-  a  d 

tude  is  5.     What  is  the  area  of  A  ABC  ? 

Suggestion.  —  Draw  AD  \\  EC  and  CD  \\  AB  to  form 
O  ABCD.  Compare  A  ABC  with  EJ  ABCD.  Then 
determine  the  area  of  O  ABCD  and  finally  ot  AAB  G. 


Propositiq]^  v.     Theorem 
333.   The  area  of  a  triangle  equals  one  half  the  prod- 


uct of  its  base  and  altitude. 


Hypothesis.     A  ABC  has  altitude  AE  =  a  and 
Conclusion.  Area  of  A  ABC=  ^  ah. 

[Proof  to  be  given  by  the  pupil.] 
Suggestion. — Construct  CJ  ABCD  and  proceed  as  in  Ex.  25. 


BC  =  h. 


AREAS   OF   POLYGONS  199 

334.  Corollaries.  By  proofs  similar  to  those  given  in  §  332, 
it  follows  that :  , 

(1)  Triangles  having  equal  bases  and  equal  altitudes  are^equal. 

(2)  Tioo  triangles  are  to  each  other  as  the  products  of  their 
bases  by  their  altitudes. 

(3)  Triangles  having  equal  altitudes  are  to  each  other  as  their 
bases. 

(4)  Triangles  having  equal  bases  are  to  each  other  as  their 
altitudes. 

(5)  A  triangle  is  one  half  a  parallelogram  having  the  same  base 
and  altitude. 

Ex.26,     (a)  CompsLTeCJ  ABCD  with  A  BCE.  a      x        d 

(6)  Compare  A  BOX  with  A  BCE. 
(c)  If  Xis  the  mid-point  of  AD,  compare  A  ABX 

with  A  XCZ>;  also  compare  A  XCD  with  A  5 O-E. 

Ex.  27.  Determine  the  area  of  an  isosceles  right  triangle  whose  leg 
is  9  in. 

Ex.  28.  Determine  the  area  of  an  equilateral  triangle  whose  side  is 
10  in. 

Ex.  29.     (a)  Prove  that  the  area  of  an  equilateral  triangle  whose 

side  is  s  is  —  v3.     (Memorize  this  formula.) 
4 
(&)  Using  the  formula  developed  in  (a),  obtain  the  area  of  an  equi- 
lateral triangle  whose  side  is:  (1)  12  in.  ;  (2)  15  in. 

Ex.  30.  Find  the  area  of  the  front  of  the  garage  whose  K('y\  lo' 
dimensions  are  indicated  in  the  adjoining  figure. 

Ex.  31.     What  is  the  area  of  the  rhombus  whose  diago- 
nals are  10  and  16  respectively  ? 

Ex.  32.  What  is  the  length  of  the  side  of  a  square  whose  area  equals 
that  of  a  triangle  whose  base  is  24  and  whose  altitude  is  12  ? 

Ex.  33.  If  BD  is  the  median  to  side  AC  of  A  ABC,  prove  that 
A  ABD  =  ABDC.     (Draw  the  altitude  BF  to  side  AC. ) 

Ex.  34.  Prove  that  the  diagonals  of  a  parallelogram  divide  the  par- 
allelogram into  four  equal  triangles. 

Ex.  35.  If  segments  are  drawn  from  two  opposite  vertices  of  a  quad- 
rilateral to  the  mid-point  of  the  diagonal  joining  the  other  two  vertices, 
the  broken  line  so  formed  divides  the  quadrilateral  into  two  equal  parts. 


200  PLANE    GEOMETRY  — BOOK   IV 

Ex.  36.  Through  the  vertex  ^  of  A  ABC,  draw  a  line  MN  parallel 
to  BC.     On  MN^  take  any  point  X  and  prove  that  A  XBC  =  A  ABC. 

Ex.  37.     Construct  a  triangle  twice  as  large  as  a  given  triangle  : 

(a)  having  the  same  base  as  the  given  triangle  ; 

(&)  having  the  same  altitude  as  the  given  triangle. 

Ex.  38.     Construct  a  rectangle  equal  to  a  given  triangle. 

Ex.  39.     Construct  a  triangle  equal  to  a  given  rectangle. 

Ex.  40.  Construct  a  right  triangle  equal  to  a  given  triangle  and  hav- 
ing the  same  base  as  the  triangle. 

Ex.  41.  Construct  an  isosceles  triangle  equal  to  a  given  triangle  and 
having  the  same  base  as  the  given  triangle. 

335.  The  Area  of  a  Triangle  can  be  expressed  in  Terms  of 
its  Sides. 

Solution.  1.  If  a,  5,  and  carethe  sidesof  A^iiC,  ands  =  ^(a  +  &  +  c), 
it  can  be  proved  that  the  altitude  drawn  to  side  a  is  given  by  the  formula  : 


ha=-Vs{s-a){s-b){s-c)  §313 

a 

Note. — The  proof  may  be  read  if  desired.     Often  in  mathematics,  we 
use  provable  formulae  which  we  may  not  have  proved  ourselves. 

2.  Areaof  A^^C  =  ia- /i„. 

3.  .-.  areaof  A^i?(7  =  -  a-  -  >/s(s  -  a)(s-  b)(s-c), 

2        a 


4.    or  area  of  A ^j5C  =  vs(s  — a)(s— 6) (s  —  c). 

Eemember  that  s  is  one  half  the  perimeter  of  the  triangle. 
Example.  —  Find  the  area  of  the  triangle  whose  sides  are 
13,  14,  and  15. 

Solution.  1.    Let  a  =  IS,  6  =  14,  and  c  =  15. 

2.  .-.  8  =  1(13  +  14+15)  =21. 

3.  .  •.  area  =  V2I  x  8  x  7  xli  ■=  V3x7x2x4x7x3x2 

4.  •  =  3  X  7  x  2  X  2  =  84. 


Ex.  42.  The  sides  of  the  lots  A 
and  B  in  the  adjoining  figure  have  the 
lengths  indicated.  Find  the  area  of 
each  of  the  lots. 


Note.  —  Supplementary  Exercises  6  to  30,  p.  295,  can  be  studied  now. 


AREAS   OF   POLYGONS  201 

Proposition  VI.     Problem 
336.   Construct  a  triangle  equal  to  a  given  polygon. 


G  D 
Given  polygon  ABODE. 
Required  to  construct  a  A  =  ABODE. 

1.  Change  ABODE  into  an  equal  quadrilateral. 
Construction.     1.   Draw  diagonal  AO,  cutting  off  A  ABO. 

2.  Draw  BF II  AO^  meeting  DC  extended  at  F.     Draw  AF. 
Statement.  AFDE  =  ABODE. 

Proof.  1.  A  ABO  and  A  AOF  have  the  same  base,  AO, 
and  equal  altitudes,  —  the  distance  between  the  lis  AO  and  BF. 

2.  .:  AABO=^AOF.  Why? 

3.  AFDE  =  AODE  +  A  AOF-, 

and  ABODE  =  AODE  -h  A  ABO.  Ax.  3,  §  51 

4.  .-.  AFDE  =  ABODE.  Ax.  7,  §  51 
II.     OJiange  AFDE  into  an  equal  triangle. 

Construction.  1.  Draw  AD ;  draw  OE  II  ADy  meeting  FD 
extended  at  6?;  draw  AG, 

Statement.     A  AFG  =  quadrilateral  AFDE.    Prove  it. 

Ex.  43.  (a)  Make  a  reasonably  large  pentagon,  and  construct  a  tri- 
angle equal  to  the  pentagon.  (6)  Measure  the  base  and  altitude  of  the 
triangle,  and  compute  the  area  of  the  triangle,  (c)  What  is  the  area  of 
the  pentagon  ?  D  15        C     -e 

Ex.  44.     Determine  the  area  of  the  trapezoid       /j     ""^-^  \    J 

ABCD  whose  dimensions  are  indicated  in  the  ad-      /l'^  ^"^^     \! 

joining  figure. 


A  F  B 

Suggestions.  —  Area  of  A  ABD  =  ?    Area  of  A  BCD  =  ? 


202 


PLANE    GEOMETRY  —  BOOK   IV 


Proposition^  VII.     Theorem 
337.   Tlie  area  of  a  trapezoid  equals  one  half  its  alti- 


tude multiplied  hy  the  sum  of  its  bases. 


A 


D         b'            0 

1 

\i 

B 


Hypothesis.     Trapezoid  ABCD  has  its  altitude  DE  =  a,  its 
base  AB  =  h,  and  its  base  CD  =  b\ 

Conclusion.  Area  ABCD  =  ^a(b  +  b'). 

Proof.     1.   Draw  BD  and  altitude  BF  of  A  DBC. 
Complete  the  proof.     See  Ex.  44. 

338.   Cor.     Tlie  area  of  a  trapezoid  equals  the  product  of  its 
altitude  and  its  median.     (Recall  §  153.) 

Ex.  45.     Determine  the  area  of  the  trapezoids  A  and  B  whose  dimen- 
sions are  given  in  the  table  below : 


Altitude 

Lower  Hase 

Upper  Base 

Trapezoid  A 
Trapezoid  B 

10  in. 

20  in. 

9  in. 

15  ft. 

30  ft. 

20  ft. 

Ex.  46.     Find  the  lower  base  of  a  trapezoid  whose  area  is  675  sq.  ft., 
upper  base  35  ft.,  and  altitude  15  ft. 

Ex.  47=  The  non-parallel  sides,  AB  and  CD, 
of  a  trapezoid  are  each  25  in.,  and  the  sides  AD 
and  BC  are  33  in.  and  19  in.,  respectively.  Find  the 
area  of  the  trapezoid.  A 

Suggestions.  —  Draw  through  jB  a  1|  to  CD,  and  al 
to  AD. 

Ex.  48.     Prove  that  the  straight  line  joining  the  mid-points  of  the 
bases  of  a  trapezoid  divides  the  trapezoid  into  two  equal  trapezoids. 


AREAS  OF  POLYGONS 


203 


Ex.  49.  How  many  square  feet  of  wood  will  be  re- 
quired for  100  waste-paper  boxes  like  the  one  pictured  in 
the  adjoining  figure,  —  allowing  15%  extra  for  wood 
wasted  in  cutting  ? 

Note.  —  Assume  that  each  side  is  an  isosceles  trapezoid 
having  the  dimensions  indicated  in  the  figure. 

Ex.  50.  The  adjoining  figure  represents  the  end  of 
a  barn.  If  the  barn  is  85  ft.  in  length,  determine  the 
expense  of  painting  its  sides,  its  end,  and  its  roof  at  4  j^ 
per  square  foot. 


Ex.  51.  The  longest  diagonal  AD  of  pentagon 
ABODE  is  44  in.,  and  the  perpendiculars  to  it  from 
B,  C,  and  E  are  24,  16,  and  15  in.  respectively. 
If  AB  =  25  in.  and  CZ)  =  30  in.,  what  is  the  area 
of  the  pentagon  ? 


Ex.  52.  Find  the  lower  base  of  a  trapezoid  whose  area  is  9408  sq. 
ft.,  whose  upper  base  is  79  ft.,  and  whose  altitude  is  96  ft. 

Ex.  53.  Construct  a  triangle  equal  to  a  given  trapezoid  and  having 
the  same  altitude  as  the  trapezoid. 

Ex.  54.  Draw  through  a  given  point  in  one  side  of  a  parallelogram 
a  straight  line,  dividing  the  parallelogram  into  two  equal  parts. 

Ex.  55.  Construct  a  parallelogram  equal  to  a  given  trapezoid,  having 
the  same  altitude  as  the  trapezoid. 

Ex.  56.  If  AD  is  the  median  to  side  BC  of  A  ABC  and  E  is  the 
mid-point  of  AD,  then  A  BEC  =  ^  A  ABC. 

Ex.  57.  If  E  and  F  are  the  mid-points  of  sides  AB  and  AC  respec- 
tively of  A  ABC,  and  D  is  any  point  in  side  BC,  prove  quadrilateral 
AEDF=^AABC. 

Ex  58.  If  E  is  any  point  in  side  BC  oi  CJ  ABCD,  then  A  ABE  -H 
AECD^^/DABCD. 

Ex.  59.  Draw  a  straight  line  perpendicular  to  the  bases  of  a  trape- 
zoid which  will  divide  the  trapezoid  into  two  equal  parts. 

339.  The  following  Proposition  is  an  alternative  demon- 
stration of  the  Pythagorean  Theorem  given  in  Proposition 
XIV  of  Book  III. 


204 


PLANE    GEOMETRY  —  BOOK   IV 


Proposition  VIII.     Theorem 

340.  The  square  upon  the  hypotenuse  of  a  right  tri- 
angle is  equal  to  the  sum  of  the  squares  upon  the  two 
legs  of  the  triangle. 


Hypothesis.     Z  C  of  A  ABC  is  a  right  angle. 

ABEF,  ACGB,  and  BGKL  are  squares. 
Conclusion.     Area  ABEF  =  area  ACGH+  area  BCKL. 
Proof.   1.     Draw  CD  ±  AB  and  extend  it  to  meet  FE  at  M. 

2.  Draw  BH  and  CF. 

3.  •  A  ACF  ^  A  ABH.     (Give  the  full  proof.) 

4.  BCO  is  a  st.  line  and  parallel  to  AH.  §  40. 

5.  .-.A  ABH  and  D  ACGH  have  the  same  base,  AH,  and 
eqiial  altitudes,  —  the  distance  between  the  lis  BO  and  AH. 

6.  .-.  area  ACGH  =  2  area  A  ABH.  Why  ? 

7.  Similarly  area  ^Z)3fi^=  2  area  A  ACF. 

(Give  the  full  proof.) 

8.  .-.  area  ACGH  =  area  ADMF.  Why  ? 

[From  steps  6  and  7.] 

9.  Similarly  it  can  be  proved  that 

area  BCKL  =  area  BDME. 
10.  .-.  area  ACGH  +  area  BCKL  =  area  ABEF. 

[From  steps  8  and  9.] 


AREAS   OF  POLYGONS 


205 


Note.  —  Many  other  proofs  of  this  important  theorem  can  be  given. 
The  proof  suggested  in  Ex.  00,  which  follows,  is  an  extremely  suggestive 
one  ;  the  one  in  Ex.  61  has  of  course  special  interest. i 

Ex.  60.  Prove  the  Pythagorean  Theorem,  us- 
ing the  adjoining  figure.  (Note.  Square  AH  is 
"  turned  in  "  over  A  ABC.) 

Prove  [JAD  =  UBF-\-U  AH. 

Suggestions.  —  1.  Draw  J5^X"  and  prove  HKE  is  a 
St.  line,  by  proving  LAKE  is  a  rt.  Z. 

2.  Prove  [J  AH  =  CJ  AXYE,  by  comparing  each 
with  A  ABE. 

3.  FroyeOBF  =  CJCXYD. 
Note.  —  The  Pythagorean  Theorem  can  be  proved 

from  figures  obtained  by  "  turning  in  "  any  of  the  squares,  one  at  a  time,  two 
at  a  time,  or  all  three  of  them. 


~~^E 


Ex.  61.     Garfield's  Proof  of  the  Pjrthagorean      a 
Theorem. 

Hyp.     In  AABC,ZB  =90^". 
Con.  b'  =  (fi  +  c2. 

Suggestions. —  1.  Extend  BC  to  D,  making  CD  = 
An.  Draw  DEIBD  at  D,  making  DE  =  BC. 
Draw  CE  and  AE. 

2.  Prove  ABDE  is  a  trapezoid. 

3.  Express  the  area  of  ABDE  in  terms  of  a  and  c. 

4.  Prove  £2  =  90°,  and  that  CE  =  b. 

5.  Express  the  area  of  A  ABC,  CDE,  ACE  in  terms  of  a,  b,  and  c. 

6.  Form  an  equation  based  on  the  fact  that  the  trapezoid  consists  of  the 
three  triangles. 

7.  Complete  the  proof  algebraically.' 

Ex.  62.     Prove  C,  H,  and  L  lie  in  a  st.  line.     (Fig.  §  340.) 
(Draw  CH  &nd  CL,  and  prove  Z  HCL  =  1  st.  Z.) 

Ex.  63.     Prove  AG  ||  BK. 

Ex.  64.  Prove  that  the  sum  of  the  Js  from  H  and  L  to  AB  extended 
equals  AB. 

Suggestion.  —  Compare  AD  and  DB  with  Jfe. 

1  A  number  of  alternative  proofs  of  the  Pythagorean  Theorem,  and  other 
interesting  theorems,  are  given  in  Heath's  Mathematical  Monographs,  Num- 
bers 1-4.  Published  by  D.  C.  Heath  &  Co.,  Boston,  New  York,  Chicago.  10/* 
each. 


206 


PLANE    GEOMETRY  —  BOOK   IV 


Proposition  IX.     Problem 

341.   Construct  a   square  equal    to  the  sum  of  two 
given  squares. 


N 


A  B 

Given  squares  M  and  N. 

Required  to  construct  a  square  equal  to  the  sum  of  M  and  N. 
Construction.     1.    Construct  ^(7  _L^-B,  making  ^C=  n  and 
AB  =  m.     Draw  BC. 

Statement.     The  square  constructed  on  J50  as  side  ==M-{-N. 
[Proof  to  be  given  by  tlie  pupil.] 

342.    Cor.     Construct  a  square  equal  to  the  difference  between 
two  given  squares. 


A  yC      D 

Given  squares  M  and  N. 

Required  to  construct  a  square  equal  to  M—  N. 
Construction  and  proof  to  be  given  by  the  pupil. 
[Construction  suggested  by  the  figure.] 

Ex.  65.     Construct  a  square  equal  to  the  sum  of  three  given  squares. 

Ex.  66.     The  area  of  an  isosceles  right  triangle  is  equal  to  one  fourth 
the  area  of  the  square  described  upon  the  hypotenuse. 

Suggestion.— Com^2iYe  the  right  triangle  with  the  square  on  one  leg. 

Ex.  67.     In  the  figure  for  §  340,  prove  that  A  AFH,  BEL,  and  CGK 
each  equals  A  ABC. 


AREAS  OF   POLYGONS 


207 


Proposition  X.     Theorem 

343.   The  areas  of  two  similar  triangles  are  to  each 
other  as  the  squares  of  any-  two  homologous  sides. 


Hypothesis.     AB  and  A^B'  are  homologous  sides  of  similar 
A  ABC  and  A'B'C  respectively. 


Conclusion. 


A  ABC       AB' 


AA'B'C     JT^i'' 

Proof.     1.   Draw  altitudes  CD  and  CD'. 

A  ABC         lAB'CD  AB -  CD 


AA'B'C 


^AB'CD 
.  A'B' '  CD' 


A  ABC  ^(  ^B\ 
AA'B'C     \A'B'J 

But 
.     A  ABC 


CD\ 

CD')' 

CD 


A'B' .  CD' 

§  333;  Ax.  6,  §  51 

An  algebraic  change 


AB 


CD'     A'B'  

AB       AB       AB^ 


A  A'B'C     A'B'     A'B' 


§282 
Ax.  2,  §  51 


A'B'' 

Note.  —  Since  the  ratio  of  two  homologous  lines  of  two  similar  tri- 
angles equals  the  ratio  of  any  two  homologous  sides,  the  areas  of  two 
similar  triangles  are  to  each  other  as  the  squares  of  any  two  homologous 
lines. 

Ex.  68.    A  ABC  ~  A  A'B'C  and  AB  =  2  A'B'. 

(a)  Compare  the  area  of  A  ABC  with  the  area  of  A  A'B'C. 

(b)  Draw  a  figure  to  illustrate  the  correctness  of  your  result. 

Ex.  69.  What  is  the  ratio  of  A  ABC  to  A  A'B'C,  if  they  are  similar, 
and: 

(a)  itAB  =  S  A'B'  ?     (b)  if  AB  =  A'B'  ?     (c)  it  AB  =  ^  A'B'  ? 

Note.  —  Supplementary  Exercises  32  to  35,  p.  297,  can  be  studied  now. 


208 


PLANE    GEOMETRY  —  BOOK   IV 


Proposition  XI.     Theorem 

344.    The  areas  of  two  similar  polygons  are  to  each 
other  as  the  squares  of  any  two  homologous  sides. 


Hypothesis.     AB  and  A'B'  are  homologous  sides  of  similar 
polygons  AC  and  A'C. 

Conelusion.  Area  of  polygon  AO  ^  AW^ 

Area  of  polygon  A'C      A'B'^ 
Proof.     1.    Draw  the  diagonals  EB,  EC,  E'B',  and  E'C. 
2.    Then  A  ABE  ~  A  A'B'E' ;  A  BCE  ^  A  B'G'E' ;  etc. 

_  §295 

.     A  ABE        AB" 


A  A'B'E' 
A  BCE 


A'B'' 
^BCf  ^AB" 

AB'C'E'      B^'""     A'W 

[Since   AB__BC_^ 
L  A'B'~  B'C  J 


3. 

4.  Similarly 

5.  Similarly 

6. 

A  A'B'E'      A  B'OE'      A  C'D'E' 

Complete  the  proof,  applying  §  296. 


§343 


ACDE 


CD" 


A  C'B'E'      c'j)'^ 
A  ABE        A  BCE 


AW 

JJb'^ 

A  CDE 


Why 


345.  Since  the  perimeters  of  two  similar  polygons  have  the 
same  ratio  as  any  two  homologous  sides  (§  297),  then  the  areas 
of  two  similar  polygons  must  have  the  same  ratio  as  the  squares 
of  their  perimeters. 


AREAS   OF   POLYGONS 


209 


Proposition  XII.     Theorem 

346.  Two  triangles  having  an  angle  of  one  equal  to 
an  angle  of  the  other  are  to  each  other  as  the  jproducts 
of  the  sides  including  these  angles. 


Hypothesis.     A  ^IBC  and  A  AB'C  have  Z  A  common. 

Conclusion.  AABO^ABxAC^ 

A  AB'C     AB'xAC 

Proof.     1.   Draw  B'C;  also  draw  CD±AB'. 

2.  A  ABC  and  A  AB'C  have  the  common  altitude  CD. 

3.  .    AABC^AB^^  Cor.  3,  §334 

A  AB'C     AB' 

4.  A  AB'C  and  A  AB'C  have  as  common  altitude  the  ± 
from  B'  to  AC. 

5.  ,AAB!C^AC  ^, 

AC  ^ 


A  AB'C 

6.   Multiplying  the  equations  of  steps  3  and  5, 
A  ABC      A  AB'C  ^  AB  x  AC         A  ABC  ^  AB  x  AC 
A  AB'C  '  A  AB'C     AB'  x  AC'  °^  A  AB'C     AB'  x  AC' 

Ex.  70.  If  the  area  of  a  poly^fon,  one  of  whose  sides  is  16  in.,  is  375 
sq.  in.,  what  is  the  area  of  a  similar  polygon  whose  homologous  side  is 
18  in.  ? 

Ex.  71.  The  longest  sides  of  two  similar  polygons  are  18  and  3  in. 
respectively.  How  many  polygons,  each  equal  to  the  second,  will  form  a 
polygon  equal  to  the  first  ? 

Note.  — Supplementary  Exercises  36  to  41,  p.  298,  can  be  studied  now. 


210 


PLANE    GEOMETRY  —  BOOK   IV 


SUPPLEMENTARY  TOPICS 

Three  groups  of  supplementary  material  follow.  This 
material  appears  in  some  form  in  most  geometries.  All  of  it 
is  interesting  and  instructive  mathematically;  none  of  it  is 
strictly  necessary  in  subsequent  parts  of  geometry. 

The  teacher  should  feel  free  to  select  the  group  or  groups 
which  best  meet  the  needs  of  the  class. 

Group  A. — Constructions  based  upon  Algebraic  Analysis. 
This  group  is  especially  instructive  and  interesting. 

Group  B.  —  Constructions  without  Formal  Analysis. 
Group  C. — Miscellaneous  Problems. 

The  first  two  problems  of  this  group  are  usually  studied.  Teachers 
often  omit  the  remaining  ones. 


A.  Constructions  based  on  Algebraic  Analysis 
Proposition  XIII.     Problem 
347.    Construct  a  square  equal  to  a  given  parallelogram. 


1 
1 

1 

s 

—Ir ^ 


Given  O  ABCD  having  base  h  and  altitude  a. 
Required  to  construct  a  square  equal  to  O  ABCD. 

Analysis.     1.   Let  x  —  the  side  of  the  required  square. 

2.  Then  x^  =  the  area  of  the  required  square,  Why  ? 
and             ah  =  the  area  of  the  given  parallelogram.        Why  ? 

3.  .'.x'^ah.  Why? 

4.  .'.a:x  =  x:h.  §  252 

5.  .-.  X  is  the  mean  proportional  between  a  and  6,  and  can 
be  constructed  by  §  290. 


CONSTRUCTIONS  2 1 1 

Construction.  1.  Construct  a;,  the  mean  proportional  between 
a  and  b.  §  290 

2.  On  X  as  side,  construct  the  square  S, 

Statement.  Square  S  =  CJ  ABCD. 

Proof.     1.    Area  of  S  =  x^y  and  area  of  O  ABCD  =  ab. 

2.  a  :  X  =  X  :  by  OT  x^  =  ab.  Why  ? 

3.  .-.  area  of  /S  =  area  of  O  ABCD, 

Discussion.  The  construction  is  always  possible,  for  the 
mean  proportional  between  a  and  b  can  always  be  found. 

Note  1.  — The  analysis  gives  the  pupil  an  idea  of  how  such  a  construc- 
tion is  discovered.  In  many  cases  the  proof  of  the  correctness  of  the 
resuking  construction  is  ratlier  trivial  after  such  an  algebraic  analysis,  — 
and  in  such  cases  the  teacher  may  decide  to  omit  the  proof. 

Note  2.  —  The  algebraic  solution  of  such  a  problem  as  that  proposed 
in  §347  would  duplicate  the  analysis  as  far  as  step  3.  Then  the  4th 
step  would  be:  .-.  x  =Vab,  After  x  had  been  computed,  the  square 
would  be  constructed  upon  a  line  of  the  length  determined. 

Theoretically  the  geometric  solution  is  preferable,  for  x  as  constructed 
actually  equals  the  mean  proportional  between  a  and  6,  so  that  the 
square  on  side  x  actually  equals  the  parallelogram  ;  whereas,  in  the  case 
of  the  algebraic  solution,  the  value  of  x  is  determined  only  approximately 
(in  most  cases)  when  the  square  root  is  found,  and  hence  the  square  will 
be  only  approximately  equal  to  the  parallelogram. 

348.   Cor.    Construct  a  square  equal  to  a  given  triangle. 

Analysis.  1.  Let  x  =  the  side  of  the  required  square,  b  =  the  base  of 
the  given  triangle,  and  h  =  the  altitude  of  the  triangle. 

2.  .•.x^=ihb. 

3.  .'.  \  b  :  X  =  X  .  h,  or  X  is  the  mean  proportional  between  ^  b  and  h. 
Construction  to  be  given  by  the  pupil. 

Ex.  72.  Construct  a  square  equal  to  twice  a  given  triangle. 
Ex.  73.  Construct  a  square  equal  to  twice  a  given  square. 
Ex.  74.  Construct  a  square  which  will  be  twice  a  given  parallelogram. 
Ex.  75.  Construct  a  square  which  will  be  three  times  a  given  triangle. 
Ex.  76.  Construct  a  square  which  will  be  two  thirds  a  given  rectangle. 
Ex.  77.  Construct  a  square  which  will  be  equal  to  a  given  pentagon. 
(First  construct  a  triangle  equal  to  the  pentagon,  and  then  a  square 
equal  to  the  triangle.) 


212  PLANE    GEOMETRY  —  BOOK   IV 

Ex.  78.  Construct  a  parallelogram  which  will  equal  a  given  rectangle 
and  have  a  given  segment  as  base. 

Analysis.  1.  Let  a  —  the  altitude  and  h  =  the  base  of  the  given  rec- 
tangle, and  let  c  =  the  given  base  of  the  parallelogram.  Let  x  —  the  re- 
quired altitude  of  the  parallelogram. 

2.  Then  ex  =  ah.  Why  ? 

3.  .'.  c  :  a  =  &  :  X,  or  X  is  the  4th  proportional  to  c,  a,  and  6.        Why  ? 
Construction  left  to  the  pupil. 

Suggestion. — Construct  the  fourth  proportional  x  and  then  construct  the  ZZ7. 

Ex.  79.  Construct  a  rectangle  equal  to  a  given  rectangle,  having  a 
given  segment  as  base.     (Analyze  as  in  Ex.  78.) 

Ex.  80.  Construct  a  triangle  equal  to  a  given  triangle,  having  a  given 
segment  as  base. 

Suggestion.  —  Determine  the  altitude  as  in  Ex.  78,  then  construct  the  triangle. 
How  many  such  triangles  can  be  constructed  ? 

Ex.  81.  Construct  a  line  parallel  to  the  base  of  a  triangle  dividing 
the  triangle  into  two  equal  parts. 

Analysis.     1.    Assume  5' C  to  be  the  required  line  :  ht 

let  AB'  =x.  /  ^\ 

2.     ,^AABC^2^^^AABC^AB\  ,3      b/ \c' 

Complete  the  analysis  and  then  make  the  construction.  That  is,  de- 
termine X  first  and  then  draw  B'C  at  the  distance  x  from  A  on  AB. 

Ex.  82.  Construct  a  rectangle  having  a  given  base  and  equal  to  f 
a  given  square.     (Analyze  as  in  Ex.  78.) 

Ex.  83.  Construct  a  triangle  having  a  given  base  and  equal  to  a 
given  parallelogram. 

Ex.  84.  Construct  a  parallelogram  having  a  given  altitude  and  equal 
to  a  given  triangle. 

Ex.  85.  Construct  a  parallelogram  having  a  given  altitude  and  equal 
to  a  given  square. 

Ex.  86.  Construct  a  parallelogram  having  a  given  altitude  and  equal 
to  a  given  trapezoid. 

Ex.  87.  —  Construct  a  triangle  having  a  given  altitude  and  equal  to  a 
given  trapezoid. 

Ex.  88.     Construct  a  square  equal  to  a  given  trapezoid. 


CONSTRUCTIONS  213 

B.   Construction  without  Formal  Analysis 

349.   Clearly,  if  MN\^  a  line  parallel  a                X 

to  BO  through  Aj  and  X  is  any  point  /^^SxT' '\ 

on  MN,  then  A  XBC  =  A  ABC.  /-""^'^^^^V 

In  fact,  it  is  clear  that : 

The  locus  of  the  vertex  of  a  triangle  equal  to  A  ABC  and  hav- 
ing the  base  BC  is  a  pair  of  lines  parallel  to  BC  at  the  distance 
of  A  from  BC. 

This  fact  aids  in  making  numerous  constructions. 

Ex.  89.     (a)  Construct  a  triangle  equal  to  a  given  triangle,  having 
the  same  base  BC  but  having  the  Z  XBC  =  60^. 
(6)  Make  a  similar  construction  if  Z  XBC  =  45°. 
(c)  Make  a  similar  construction  if  Z  XBC  =  30°. 

Ex.  90.  Construct  a  A  XBC  equal  to  a  given  A  ABC,  having  tlie 
same  base  BC  and  side  XB  equal  to  a  given  segment  d. 

Ex.  91.  Construct  a  A  XBC  equal  to  a  given  A  ABC,  and  having 
the  median  from  Xto  BC  equal  to  a  given  segment  m. 

Ex.  92.  Construct  a  parallelogram  XBCYeqnaA  to  a  given  O  ABCD^ 
having  the  same  base  BC : 

(a)  having  Z  XBC  =  a  given  angle  ; 
(6)  having  side  XB  =  a  given  segment ; 
(c)  having  diagonal  YB  =  a  given  segment. 

Ex.  93.  Construct  a  triangle  equal  to  a  given  triangle  and  having 
two  of  its  sides  equal  to  given  segments  m  and  n. 

Suggestion.  —  Select  m  as  base,  and  determine  the  altitude  to  m  as  in 
Ex.  78.    Continue  as  in  §  349. 

Ex.  94.  Construct  a  triangle  equal  to  a  given  triangle  and  having  one 
side  equal  to  a  given  segment  m,  and  one  angle  adjacent  to  that  side 
equal  to  a  given  angle,  Z  T. 

Ex.  95.  Construct  a  triangle  equal  to  a  given  square,  having  given 
its  base  and  an  angle  adjacent  to  the  base. 

Ex.  96.  Construct  a  triangle  equal  to  a  given  square,  having  given 
its  base  and  the  median  to  the  base. 

Ex.  97.  Construct  a  rhombus  equal  to  a  given  parallelogram,  having 
one  of  its  diagonals  coincident  with  a  diagonal  of  the  parallelogram. 


214  PLANE    GEOMETRY  —  BOOK   IV 

C.   Miscellaneous  Supplementary  Problems 

Proposition  XIV.     Problem 

350.    Construct   a  rectangle   equal  to  a   given  square,  having 
the  sum  of  its  base  and  altitude  equal  to  a  given  segment. 


C      D,. ~ 

-"F 


M 


N 


A        E 


Given  square  M  and  segment  AB. 

Required  to  construct  a  rectangle  =  M,  having  the  sum  of 
its  base  and  altitude  =  AB. 

Construction.  1.  On  AB  as  diameter  construct  semicircle 
ADB. 

2.  Draw  AC  1.AB,  making  AC  =  side  of  M. 

3.  Draw  CF  il  AB,  intersecting  arc  ADB  at  D. 

4.  Draw  DE  ±  AB. 

5.  Construct  □  N,  having  its  base  =  BE  and  its  altitude 
=  AE. 

Statement.  Rectangle  N=  square  M. 

Proof.     1.  AE:DE  =  DE:  BE.  §  289 

2.  .-.  DE^  =  AEx  BE.  Why  ? 

3.  .-.  area  oi  M=  area  of  N.  Why  ? 
Discussion.     The  construction  is  impossible  when  the  side 

of  the  square  is  more  than  i  AB.  Why  ? 

Note.  —  §  350  suggests  a  geometrical  solution  of  a  quadratic  of  the  form 
x'^-tx  +  m^  =  0. 

From  this  equation,  7/1"^  =  x(t  —  x).  Clearly,  w  corresponds  to  a  side 
of  the  square,  x(t  —  x)  corresponds  to  the  area  of  the  rectangle  equal  to 
the  square,  and  t  corresponds  to  the  given  segment,  for  x  +  (t  —  x)  =^  t. 

Solve  ic^  —  10  X  +  16  =  0  geometrically  and  check  the  solution  alge- 
braically. 


CONSTRUCTIONS  215 


Proposition  XV.     Theorem 

351.    Construct  a  rectangle  equal  to  a  given  square,  having  the 
difference  between  its  base  and  altitude  equal  to  a  given  segment. 


kC 

M 

i    ')p, ..^ 

A\           \           1 

D 


N 


■ E 

Given  square  M  and  segment  AB. 

Required  to  construct  a  rectangle  equal  to  Jf,  having  the 
difference  between  its  base  and  altitude  equal  to  AB. 

Constmction.     1.   On  AB  as  diameter,  construct  O  ADB. 

2.  Draw  ACl.  AB,  making  AC=2i  side  of  M. 

3.  Through  the  center  0,  draw  CO,  intersecting  the  O  at 
D  and  E. 

4.  Construct  □  N,  having  its  base  =  CE  and  its  altitude 
=  CD. 

Statement.  □  -A^  is  the  required  rectangle. 

Proof.     1.  CE-CD  =  DE  =  AB', 

that  is,  the  base  of  ^—  the  altitude  of  -^=  AB. 

2.  AC  is  tangent  to  the  circle.  Why  ? 

3.  .-.  CExCD=  C7l\  §  287 

4.  .-.  area  of  ^=  area  of  M,  Why  ? 

Discussion.  The  construction  is  always  possible,  since  a 
secant  can  always  be  drawn  through  the  center  of  the  circle 
from  an  exterior  point. 

Note.  —  §  351  suggests  a  geometrical  solution  of  a  quadratic  of  the  form 
x^  ■}-  tx  —  m^  =  0,  for  this  equation  may  be  written,  m^  =  x(t  +  x). 

Solve  the  equation  x^  +  Sa;  —  9  =  0  geometrically,  and  check  the  solu- 
tion algebraically. 


216 


PLANE    GEOMETRY  —  BOOK   IV 


Proposition  XVI.     Problem 

352.    Construct   a   square   having   a   given   ratio   to  a  given 
square. 


R 

i 

a 

X 

^ 

S 

^a 

b 

'^>C}\ 

V 

o 

Given  square  R  and  the  segments  a  and  h. 

Required  to  construct  a  square  S  such  that  S :  B  =  a  :  b. 

Analysis.     1.    Let  x  =  one  side  of  the  required  square. 
2.  .  •.  x^ :  r'^  =  a:  b. 


3. 


4.    .-.  r  :  x  =  x 


.'.  bx'==ar\  ovx'  =  [~]xr. 


ar 


r  and 


b 
5.    Let 


Algebra 

or  x  is  the  mean  proportional  between 

Why? 


ar 
y  =  —,OT  by  =  ar. 

0 


6.    .'.b  :a  =  r  :y,  ov  y   is  the   fourth  proportional   to   b,   a, 
and  r.  Why  ? 

Construction.     1.   Construct  y  as  determined  in  step  6. 

2.  Construct  x  as  determined  in  step  4. 

3.  Construct  square  S  having  x  as  side. 


Statement. 

S:R  =  a:b. 

Proof.     1. 

S:R  =  x'':r^ 

2. 

But  x"^  =  ry,  and  2/  = 

ar  ^ 

b  ' 

Construction 

3. 

« '(f) 

R-     r'     ' 

Algebra 

4. 

,  S_ar^      l_a 
"  R       b    '  r^      b' 

CONSTRUCTIONS  217 

Proposition  XVII.     Problem 

353.    Construct  a  polygon  similar  to  a  given  polygon  and  hav- 
ing a  given  ratio  to  it. 

V 


A  B 

Given  polygon  AC  and  segments  a  and  6. 

Required  to  construct  a  polygon   A'O   similar  to  AC  and 
polygon  AC      a 

such  that  —, yrvy,  =  T  • 

polygon  A'C     o 
Analysis.     1.   Let  x  =  the  side  homologous  to  AB. 

2.   Then  yolygon  AC  ^AW_  ^  344 

polygon  A'C        x' 

o  AW     a 

Complete  the  analysis  as  in  Prop.  XVI,  thus  determining  x 
in  terms  of  AB^  a,  and  h.  Then  construct  upon  x  as  side 
homologous  to  AB  a  polygon  similar  to  polygon  AC,  by  §  294. 
This  will  be  the  required  polygon. 

Note.  —  Notice  that  §  362  is  a  special  case  of  §  363,  for  all  squares  are 
similar. 

Ex.  98.  Construct  a  rectangle  similar  to  a  given  rectangle  and  having 
to  it  the  ratio  2:1. 

Ex.  99.     Construct  an  equilateral  triangle  equal  to  a  given  triangle. 

Suggestion.  —  Determine  the  side  s  (Ex.  29,  Book  IV)  as  in  §  348.  Recall 
Ex.  81,  Book  III. 

Ex.  100.     Construct  a  triangle  equal  to  the  sum  of  two  given  triangles. 

Suggestion.  —  First  construct  squares  equal  to  the  given  triangles. 

Ex,  101.  Construct  a  triangle  equal  to  the  difference  of  tvv^o  given 
triangles. 

Ex.  102.  Draw  a  line  parallel  to  the  ba.se  of  a  triangle  which  will 
divide  the  triangle  into  two  parts  which  will  have  the  ratio  1 : 2, 

Suggestion.  —  A.xxq\jz%  as  hi  Ex.  81,  Book  IV. 


218  PLANE    GEOMETRY  —  BOOK   IV 

Proposition  XYIII.     Problem 

354.   Construct  a  polygon  similar  to  one  of  two  given  polygons 
and  equal  to  the  other. 


m 


Given  polygons  iWand  N. 

Required  to  construct  a  polygon  similar  to  M  and  equal  to  N. 

Analysis.  1.  Let  m=  the  side  of  the  square  equal  to  J/, 
and  n  =  the  side  of  the  square  equal  to  N.  Let  x=  that  side 
of  the  required  polygon  P  which  is  homologous  to  AB. 

2.  ...^=^.  Why? 

3.  But  ^,=  ~  =  —•     Since  P  =  N. 

Why 

5.  r.-  =  ^^-.  Why? 
n        X 

6.  Hence  x  is  the  fourth  proportional  to  m,  n,  and  AB. 
Construction.     1.    Construct  the  squares  equal  to  M  and  N, 

thus  determining  segments  m  and  n.  See  Ex.  77 

2.  Construct  x  as  determined  in  step  6. 

3.  Upon  X  as  side  homologous  to  AB,  construct  a  polygon 
P  similar  to  Jf.  §  294 

Statement.  P  ^  M,  and  P=N. 

Proof.     1.  P-^M.     Why? 

2.  P:  Jf=a;2:ZB2.  Why? 

3.  But  X  =  ^^^^ .  Construction  2 

m 

4.  .*.  P :  Jf  =  71^ :  ml  Substituting  in  step  2 

5.  .-.  P:  il^=  iV':  J/,  or  P  =  iV: 


■•£= 

M 

^.     Since  P 
m^ 

m^ 

X^ 

AB" 

m 

a; 
^5 

m 
n 

^5 

MISCELLANEOUS    EXERCISES  219 

Miscellaneous  Exercises 

Ex.  103.  A  road  CO  ft.  wide  runs  from  one  corner  to  the  opposite 
corner  of  a  square  field  measuring  500  ft.  on  a  side,  the  diagonal  of  the 
field  running  along  the  center  of  the  road.  What  is  the  area  of  that 
portion  of  the  field  occupied  by  the  road  ?  (Carry  out  the  results  to  two 
decimal  places.) 

Ex.  104.  What  is  the  length  of  the  side  of  an  equilateral  triangle 
equal  to  a  square  whose  side  is  15  in.? 

Suggestion.  —  Recall  Ex.  29,  Book  IV. 

Ex.  105.  From  one  vertex  of  a  parallelogram,  draw  lines  dividing  the 
parallelogram  into  three  equal  parts. 

Ex.  106.  The  sides  AB,  BC,  CD,  and  DA  of  quadrilateral  ABCD  are 
10,  17,  13,  and  20  respectively,  and  the  diagonal  ^O  is  21.  Find  the 
area  of  the  quadrilateral. 

Ex.  107.  If  diagonals  ^Cand  BD  of  trapezoid  ABCD  intersect  at  E, 
then  AAEB  =  A  DEC.    {BC  and  AD  are  the  bases  of  ABCD.) 

Suggestion.  —  Compare  A  ABD  and  A  ACD. 

Ex.  108.     If  Xis  any  point  in  diagonal  AC  of  O  ABCD,  then 
AABX=AAXD. 

Suggestion. — Draw  the  altitudes  from  B  and  D  to  base  AX. 

Ex.  109.  If  E  and  F  are  the  mid-points  of  sides  AB  and  ^C  of 
A  ABC,  then  A  AEF  ;=  J  A  ABC. 

Ex.  110.  If  E  is  any  point  within  O  ABCD,  then  A  ABE  +  A  CDE 
equals  I  the  parallelogram. 

Suggestion.  —Throngh  E  draw  a  line  parallel  to  AB. 

Ex.  111.  If  Z  ^  of  A  ABC  is  30°,  prove  that  the  area  of  A  ABC  = 
\ABx  AC. 

Suggestion.  —  Draw  BD  1  AC.    Recall  Ex.  128,  Book  I. 

Ex.  112.  Prove  that  the  area  of  a  rhombus  is  one  half  the  product  of 
its  diagonals. 

Ex.  113.  If  E  is  the  mid-point  of  CD,  one  of  the  non-parallel  sides  of 
trapezoid  ABCD,  prove  that  ABE  =  ^  ABCD. 

Suggestion.  —  Through  E,  draw  a  line  parallel  to  AB. 

Ex.  114.  Construct  an  isosceles  triangle  equal  to  a  given  triangle, 
having  given  one  side  of  length  m. 

Suggestion.  —  Use  m  as  the  base.  Determine  the  altitude  to  m  as  in  Ex.  78, 
Book  IV.    Then  follow  §  241. 


220  PLANE    GEOMETRY  —  BOOK   IV 

Ex.  115.  Draw  through  a  given  point  in  one  base  of  a  trapezoid  a 
straight  line  which  will  divide  the  trapezoid  into  two  equal  parts. 

Ex.  116.  If  the  diagonals  of  a  quadrilateral  are  perpendicular,  the 
sum  of  the  squares  on  one  pair  of  opposite  sides  of  the  quadrilateral 
equals  the  sum  of  the  squares  on  the  other  pair. 

Note.  —  Supplementary  Exercises  42  to  46,  p.  298,  can  be  studied  now. 

Review  Questions 

1.  Define  area  of  a  plane  figure. 

2.  Distinguish  between  congruent,  similar,  and  equal  figures. 

3.  State  the  rule  for  determining  the  area  of  : 

(a)  a  rectangle  ;  (c)  a  triangle  ; 

(6)  a  parallelogram  ;  (d)  a  trapezoid. 

4.  State  the  formula  for  the  area  of  any  triangle  in  terms  of  its  sides 
a,  6,  and  c,  and  the  number  s. 

What  is  the  number  s  ? 

5.  State  the  formula  for  the  area  of  an  equilateral  triangle  in  terms  of 
its  side  s. 

6.  State  the  corollaries  by  which  the  areas  of  two  rectangles  are 
compared  : 

(a)  If  the  rectangles  have  equal  altitudes. 

(&)  If  the  rectangles  have  equal  bases. 

(c)  When  no  known  relation  exists  between  the  altitudes  or  the  bases. 

7.  State  the  corresponding  corollaries  for  two  parallelograms. 

8.  State  the  corresponding  corollaries  for  two  triangles. 

9.  State  a  theorem  connecting  the  areas  of  a  triangle  and  a  parallel- 
ogram having  equal  bases  and  equal  altitudes. 

10.    State  a  theorem  connecting  the  areas  of  two  similar  polygons. 


BOOK  V 


REGULAR  POLYGONS.     MEASUREMENT 
OF  THE   CIRCLE 

355.  Review  the  definitions  given  in  §  125,  §  128,  and  §  178. 

356.  A  Regular  Polygon  is  a  polygon  which  is  both  equi- 
lateral and  equiangular. 

The  figures  below  illustrate  some  uses  of  regular  polygons : 


Two  Linoleum  Patterns 
Notice  the  regular  triangles,  hexagons,  squares,  and  octagons. 

Ex.  1.  Prove  that  the  exterior  angles  at  the  vertices  of  a  regular 
polygon  are  equal. 

Ex.  2.  What  is  the  perimeter  of  a  regular  pentagon  one  of  whose 
sides  is  7  in.  ?  of  a  regular  octagon  one  of  whose  sides  is  6  in.  ? 

Ex.  3.  In  §  154,  we  have  proved  that  the  sum  of  the  angles  of  any 
polygon  having  n  sides  is  (n  —  2)  st.  A. 

IIow  large  is  each  angle  of  a  regular  polygon  having :  (a)  3  sides  ? 
(6)  4  sides  ?  (c)  6  sides  ?  (d)  6  sides  ?  (e)  8  sides  ?  (/)  10  sides  ? 

Ex.  4.  (a)  Four  square  tile  can  be  used  to  cover  the  space  around 
a  point.     (Why?) 

(6)  In  the  shape  of  what  other  regular  polygon  can  tile  be  made  in 
order  that  the  surface  around  a  point  can  be  completely  covered  by 
using  tile  of  the  same  shape  ? 

357.   Each   angle   of  a   regular  polygon   having  n   sides  is 


Cv^) 


A.     (See  Ex.  3.) 


221 


222 


PLANE   GEOMETRY  —  BOOK  V 


Proposition  I.     Theorem 

358.   A  circle  can  he  circumscribed  about  any  regu- 
lar polygon. 


Hypothesis.     ABODE  is  a  regular  polygon. 
Conclusion.     A  circle  can  be  circumscribed  about  ABODE. 
Proof.     1.   A  O  can  be  constructed  through  A,  B,  and  O. 
Let  0  be  its  center  and  OA,  OB,  and  00  be  radii  of  it. 
2.    It  can  now  be  proved  that  this  circle  passes  through  D 
by  proving  OD  —  OA.     (Draw  OD.) 

Suggestions. —  1.  Compare  Z.  AB C and  L  B CD ;  Zl  and  Z 2 ;  then  Z 3 and  Z4. 

2.  Prove  A  AOB  ^  A  OCD,  and  then  OD  =  OA. 

3.  Hence  the  0  passes  through  D. 

3.  Similarly  the  circle  can  be  proved  to  pass  through  E. 

4.  Hence  a  O  can  be  circumscribed  about  ABODE. 

359.   Cor.  1.     A  circle  can  be  inscribed  in  any  regular  polygon. 

Proof.  1.  AB,  BO,  OD,  etc.  are  equal 
chords  of  the  circle  which  can  be  cir- 
cumscribed about  ABODE. 

2.  Hence  these  sides  are  equidistant 
from  0.  Why  ? 

3.  Hence  a  circle  can  be  drawn 
tangent  to  each  of  the  sides  of  ABODE. 

§  198 


REGULAR  POLYGONS  223 

360.  The  Center  of  a  regular  polygon  is  the  common  center 
of  the  circumscribed  and  inscribed  circles  ;  as  0. 

The  Eadius  of  a  regular  polygon  is  the  distance  from  its 
center  to  any  vertex ;  as  OA. 

The  Apothem  of  a  regular  polygon  is  the  distance  from  its 
center  to  any  side ;  as  OF. 

The  Central  Angle  of  a  regular  polygon  is  the  angle  between 
the  radii  drawn  to  the  ends  of  any  side;  as  Z  AOB, 

The  Vertex  Angle  of  a  regular  polygon  is  the  angle  between 
two  sides  of  the  polygon. 

q/»AO 

361.  Cor.  2.     The  central  angle  of  a  regular  n-gon  is  . 

n 

362.  Notation.     The  following  notation  will  be  employed : 
(a)  s^,  Sg,  or  s„  will  denote  one  side  of  a  regular  inscribed 

polygon  of  4,  6,  or  n  sides  respectively. 

(6)  a^j  ttg,  or  a„  will  denote  the  apothem  of  a  regular  in- 
scribed polygon  of  4,  6,  or  ri  sides  respectively. 

(c)  P4j  Pq,  or  p„  will  denote  the  perimeter  of  a  regular  in- 
scribed polygon  of  4,  6,  or  n  sides  respectively. 

(d)  7Ci,  k^  or  k^  will  denote  the  area  of  a  regular  inscribed 
polygon  of  4,  6,  or  n  sides  respectively. 

To  denote  the  corresponding  quantities  for  a  regular  circum- 
scribed polygon,  a  capital  letter  with  the  appropriate  subscript 
will  be  employed.     Thus, 

S^  =  one  side  of  the  regular  circumscribed  pentagon. 

A^  =  the  apothem  of  the  regular  circumscribed  pentagon. 

Pg  =  the  perimeter  of  the  regular  circumscribed  pentagon. 

/ig  =  the  area  of  the  regular  circumscribed  pentagon. 

Ex.  5.  Find  the  number  of  degrees  in  the  central  angle  and  in  the 
vertex  angle  of  a  regular  polygon  of  :  (a)  3  sides  ;  (6)  4  sides  ;  ^(c)  5  sides ; 
(d)  6  sides  ;  (e)  8  sides  ;  (/)  10  sides. 

Find  also  the  sum  of  the  central  angle  and  the  vertex  angle  in  each  case. 
Do  the  results  suggest  any  theorem  ? 

Ez.  6.  Prove  that  any  radius  of  a  regular  polygon  bisects  the  angle  to 
whose  vertex  it  is  drawn. 

Note.  —  Supplementary  Exercises  1  to  2,  p.  299,  can  be  studied  now. 


224 


PLANE    GEOMETRY  —  BOOK    V 


Pkoposition  II.     Theorem 

363.   The  area  of  a  regular  polygon  is  equal  to  one 
half  the  product  of  its  apothem  and  its  perimeter. 


A        F       B 
Hypothesis.     The  perimeter  of  regular  polygon  AC  is  p  and 

the  apothem  OF  is  r. 
Conclusion.  Area  of  ABODE  =  |  rp. 

Plan.     From  the  center  0  of  polygon  AC,  draw  the  radii 
OA,  OB,  OC,  etc.  forming  A  having  the  common  altitude  r. 
Determine  the  area  of  each  triangle  and  add  the  results. 


Proposition  III.     Theorem 

364.  If  a  circle  he  divided  into  any  number  of  equal 
arcs,  the  chords  of  these  arcs  form  a  regular  inscribed 
polygon  of  that  number  of  sides. 


Hypothesis.     AB=BC=  CD=BE^EA  in  O  0. 
Conclusion.     ABODE  is  a  regular  pentagon. 

[Proof  to  be  given  by  the  pupil.] 
Note.  —  Supplementary  Exercise  3,  p.  299,  can  be  studied 


REGULAR   POLYGONS 


225 


365.  Cor.  1.  If  from  the  mid-point  of 
each  arc  subtended  by  a  side  of  a  regular 
polygon  lines  be  drawn  to  its  extremities^  a 
regular  inscribed  polygon  of  double  the  num- 
ber of  sides  is  formed. 

366.  Cor.  2.  An  equilateral  polygon 
inscribed  in  a  circle  is  regular. 

Note.  —  Supplementary  Exercises  4  to  5,  p.  299,  can  be  studied  now. 


Proposition  IV.     Theorem 

367.  If  a  circle  is  divided  into  any  number  of  equal 
arcs,  the  tangents  at  the  points  of  division  form  a  regu- 
lar circumscribed  polygon  of  that  number  of  sides. 


Hypothesis.  O  ACD  is  divided  into  five  equal  arcs,  AB,  BCj 
etc.  X  T,  YZ,  etc.  are  tangent  to  O  ACD  at  A,  B,  etc.,  form- 
ing pentagon  XTZWK. 

Conclusion.     X  YZ  WK  is  a  regular  pentagon. 

Suggestions.     1.   Draw  ^B,  ^C,  (7Z>,  etc. 

2.  Prove       ^  AXE,  A  YB,  etc.  congruent  isosceles  A. 

3.  Prove  AX  =  AY=  BY=  BZ,  etc. 

4.  Prove  XY  =  YZ  =  ZW,  etc. 
Recall  §  356.     Complete  the  proof. 

Ex.  7.  Prove  that  any  apothem  of  a  regular  polygon  bisects  the  side 
to  which  it  is  drawn. 

Ex.  8.  Prove  that  the  diagonals  drawn  from  one  vertex  of  a  regular 
hexagon  divides  the  angle  at  the  vertex  into  4  equal  angles. 


226 


PLANE    GEOMETRY  —  BOOK  V 


Corollaries  to  Proposition.  IV 

368.  Cor.  1.  Tarigents  drawn  to  the  cir- 
cle at  the  mid-points  of  the  arcs  iiicladed  be- 
tween two  consecutive  points  of  contact  of  a 
regular  circumscribed  polygon  form,  with  the 
sides  of  the  original  circumscribed  polygon, 
a  regular  circumscribed  polygon  having 
double  the  number  of  sides. 

YoT  the  circle  is  divided  into  double  the  number  of  equal 
arcs.     The  theorem  follows  by  §  367. 

369.  Cor.  2.  Tangents  drawn  to  the 
circle  at  the  mid-points  of  the  arcs  sub- 
tended by  the  sides  of  a  regular  inscribed 
polygon  form  a  regular  circumscribed 
polygon  of  the  same  number  of  sides. 

If  ABODE  is  regular,  and  F,  G,  H, 
K,  3ind  L  are  the  mid-points  of -arcs 

AB,   BC,   etc.,  then  LF  =  FG,    etc. 
Hence  A'B'C'D'E'  is  regular. 

Note.  — Supplementary  Exercise  6,  p.  299,  can  be  studied  now. 

370.  Construction  of  Regular  Polygons  is  based  upon  Prop- 
ositions III  and  IV.  In  order  to  divide  a  circle  into  any  number 
of  equal  parts,  it  is  sufficient  to  be  able  to  divide  the  total  angle 
around  the  center  into  that  same  number  of  equal  angles. 

Ex.  9.  By  using  your  compass,  ruler,  and  protractor,  draw  a  regu- 
lar inscribed  pentagon  within  a  circle  of  radius  2.5  in. 

Ex.  10.  Draw  a  regular  polygon  of  9  sides  within  a  circle  of  2.5  in. 
as  in  Ex.  9. 

371.  It  is  customary  in  geometry,  however,  to  use  only  the 
comjjass  and  straightedge  in  making  constructions.  It  is  in- 
teresting therefore  to  inquire  what  regular  polygons  can  be 
made  by  using  only  these  tools. 


REGULAR   POLYGONS 


227 


Proposition  V.     Problem 
372.  Inscribe  a  squai^e  in  a  given  circle. 


Given  circle  0. 

Required  to  inscribe  a  square  in  circle  0. 

Construction.    1.   Draw  AG  and  BD,  perpendicular  diameters. 
2.  Draw  chords  AB,  BC,  CD,  and  AD. 

Statement.     ABCD  is  the  required  square. 

Proof.     1.  AB=BG=CD  =  DA.  Why? 

2.  .-.  ABCD  is  an  inscribed  square.  .  §  364 

373.  Cor.  1.  Regular  inscribed  x>olygons  of  8,  16,  32,  etc. 
sides  can  he  constructed.  §  365 

Note.  —  Hence,  by  §  372  and  §  373,  regular  inscribed  polygons  the 
number  of  whose  sides  is  a  number  of  the  form  2"  where  n  is  an  integer 
^  2,  can  be  constructed  by  ruler  and  compass  alone. 

Thus,  when  n  =  2,  2"  =  4  ;  when  n  =  3,  2'»  =  8  ;  etc. 

Ex.  11.  Construct  within  a  circle  having  a  3-in. 
radius  an  eight-pointed  star  like  the  one  which  forms 
the  central  unit  of  the  adjoining  linoleum  pattern. 

Ex.  12.  A  designer  wishes  to  make  a  pattern  for 
the  octagonal  top  of  a  taboret  whose  longest  diag- 
onal is  to  be  18  in.  Make  a  scale  drawing  of  the 
octagon,  letting  1  in.  represent  3  in. 

Ex.  13.  A  square  is  inscribed  in  a  circle  of  radius  10  in.  Compute  S4, 
04,  7)4,  and  kj^  to  two  decimal  places.     (See  §  362.) 

Ex.  14.  A  square  is  circumscribed  about  a  circle  of  radius  10  in. 
Compute  /S'4,  P4,  and  K^. 

Note.  — Supplementary  Exercises  7  to  14,  p.  299,  can  be  studied  now. 


228  PLANE    GEOMETRY  —  BOOK  V 

Proposition  VI.     Problem 
374.  Inscribe  a  regular  hexagon  in  a  circle, 
-^.^^ ^  C 


1^^- ^^^E 

Given  circle  0. 

Required  to  inscribe  a  regular  hexagon  in  circle  0. 

Analysis.     The  central  angle  of  a  regular  hexagon  is  60°. 

Construction.  Draw  any  radius  OA.  With  A  as  center 
and  Oxi  as  radius  draw  an  arc  cutting  the  0  at  B. 

Statement.  AB  =  \  oi  the  O,  and  may  be  applied  6  times  to 
the  circle.     The  chords  of  these  arcs  form  the  regular  inscribed 

^      '  [Proof  to  be  given  by  the  pupil.] 

Suggestions.  —  Prove  Z  AOB  =  60°  and  that  AB  =  |  of  the  circle. 

375.  Cor.  1.  Chords  joining  the  alternate  vertices  of  a  regu- 
lar inscribed  hexagon,  starting  with  any  vertex,  form  a  regular 
inscribed  triangle. 

376.  Cor.  2.  Regular  inscribed  polygons  of  12,  24,  JfS,  etc., 
sides  can  be  constructed.     (§  365.) 

Note.  — By  §§  374,  375,  and  376,  regular  inscribed  polygons  the  number 
of  w^hose  sides  is  a  number  of  the  form  3  •  2«  can  be  constructed  with  ruler 
and  compass  alone,  where  n  is  an  integer  ^  0. 

(What  is  3  .  2^  when  w  =  0?l?2?3?) 

Ex.  15.  Prove  that  the  diagonals  joining  alternate  vertices  of  a  regu- 
lar hexagon  are  equal. 

Ex.  16.  Prove  that  the  radii  of  a  regular  inscribed  hexagon  divide  it 
into  six  congruent  equilateral  triangles. 


REGULAR   POLYGONS 


229 


Ex.  17.  Prove  that  diagonals  AD,  BE,  and  CF  of  a  regular  hexagon 
ABCDEF  are  the  diameters  of  its  circumscribed  circle. 

Ex.  18.  Prove  that  the  opposite  sides  of  a  regular  hexagon  are 
parallel. 

Ex.  19.  Prove  that  the  diagonal  FC  of  regular  hexagon  ABCDEF  is 
parallel  to  sides  AB  and  DE. 

Ex.  20.  Prove  that  the  diagonal  AD  of  regular  hexagon  ABCDEF  is 
perpendicular  to  diagonal  BF  and  bisects  it. 

Ex.  21.  A  regular  hexagon  is  inscribed  in  a  circle  of  radius  10  in. 
Compute  the  lengths  of  se,  «6»  P^^  and  k^.     (See  §  362.) 

Ex.  22.  A  regular  triangle  is  inscribed  in  a  circle  of  radius  10  in. 
Compute  the  lengths  of  as,  Ss,  Pa,  and  kz. 

Ex.  23.   Find  the  area  of  a  regular  hexagon  whose  apothem  is  6  in. 

Ex.  24.   Construct  one  of  the  following  designs  : 


AB  =  6" 
CD  =  4" 


AB  =  b" 
CD  =  3" 


CD  =  6" 
EF  =  2" 


Ex.  25.  Construct  a  pattern  for  a  doily  like  the  one 
adjoining,  making  the  dimension  AB  =  9  in.  and  having 
the  radius  of  the  outer  arcs  ^  in.  longer  than  the  radius 
of  the  inner  arcs  of  the  scallops. 


Note.  —  Supplementary  Exercises  15  to  31,  p.  301,  can  be  studied  now. 

377.  A  segment  is  divided  by  a  given  point  in  Extreme  and 
Mean  Ratio  when  one  part  is  the  mean  proportional  between 
the  whole  segment  and  the  other  part. 

Thus  G  divides  AB  internally  in  extreme       ^ 
and  mean  ratio  if  C 

AB:AC=  AC:  CB. 

Notice  that  in  this  case  the  whole  segment  is  to  its  longer  part  as 
the  longer  part  is  to  the  shorter  part. 


230  PLANE    GEOMETRY  —  BOOK  V 


Proposition  YII.     Problem 

378.  Divide  a  given  segment  in  extreme  and  mean 
ratio.  ^ ^ 


A' 5 'B  A^ e^-J^ 

Fig.  1  Fig.  2 

Given  segment  AB  —  m. 

Required  to  divide  AB  in  extreme  and  mean  ratio. 

Analysis.     1.   Let  x  =  AQ  and  .•.  m  —  x=  CB. 

2.  .'.  m:x  =  x:  {m  —  x). 

3.  .*.  x^  =  m{m  —  x). 

4.  .*.  a;2  +  mx  =  m^. 


6.  ,.(.+|J  =  ^.+f^ 


2^ 

X  =w,^4-f'- 


7.    ,'.x-\ is  the  hypotenuse   of   a   right   triangle  whose 

base  is  m  and  whose  altitude  is  — . 

z 

Construction.     (Fig.  2.)   1.  Draw  EB  ±  AB,  making  AB  =  m, 
and^^  =  ^. 

7)1 

2.  Draw  ^^  and  on  it  take  EF=EB  =  ^. 

3.  On  AB,  take  yl  e  =  AF. 
Statement.  AB :  AC  =  AC :  CB. 

Note.  —  If  the  equation  of  step  6  of  the  Analysis  is  solved  for  x : 

^    4         2  2   >         2 

.•.i»  =  -(V5-l)=- (2.236-1)  =-(1.230)=. 6  m. 

2  ^  ji 


REGULAR  POLYGONS  231 

Proof.*  1.  Complete  the  circle  with  center  E  and  radius 
EB,  and  extend  AEj  cutting  the  circle  at  G. 

2.  AB  is  tangent  to  O  BFG.  Why  ? 

3.  ...  :^=:^.  §287 

AB     AF 

4.  ...  ^  =  ^.  Why? 

AB     AC  _ 

5  .  AG-AB^AB-AG  ^.^ 

"        AB  AC  ^' 

6.  But  AB=:  2  EB  =  FG.  Const. 

7.  .-.  ^G  -  yl£  =  ^(?  -  i^G^  =  AF=  AC. 
g  .   AC^CB 

' '  AB     AC 
(Substituting  in  step  5.) 

9.  ...4^  =  ^.  Why? 

AC      CB  ^ 

Note  1.  —  A  point  D  divides  a  segment  AB  externally  (see  §  305)  in 
extreme  and  mean  ratio  if  AB  :  AD  =  AD  :  DB.  This  form  of  division  is 
not  used  in  this  text. 

Note  2.  — The  Greeks  called  this  method  of  division  of  a  segment  Golden 
Section.  It  represented  to  them  the  most  artistic  division  of  a  segment 
into  two  parts. 

Ex.  26.    Find  AC  and  CB  in  §  378  algebraically  if  AB  =  10  in. 
(Let  AC  =x,  and  hence  CB  =  10  —  x.     Continue  algebraically.) 

Ex.  27.  (a)  What  is  the  relation  between  the  area  of  the  inscribed 
and  of  the  circumscribed  squares  of  a  given  circle  ? 

(6)  What  is  the  relation  between  the  perimeter  of  the  inscribed  and 
of  the  circumscribed  squares  of  a  given  circle  ? 

Ex.  28.  Prove  that  the  opposite  sides  of  a  regular  octagon  are 
parallel. 

Ex.  29.  Prove  that  diagonals  HC  and  DG  of  regular  octagon 
ABCDEFGH are  parallel . 

Ex.  30.  Prove  that  figure  ACDF  of  regular  octagon  ABCDEFGH 
is  an  isosceles  trapezoid. 

*  This  proof  may  be  omitted  if  desired. 


232  PLANE    GEOMETRY  —  BOOK   V 

Proposition  VIII.     Problem 
379.  Inscribe  a  regular  decagon  in  a  given  circle. 


Given  o  ACD. 

Required  to  inscribe  a  regular  decagon  in  O  ACD. 

Construction.     1.  Draw  any  radius  OA  and  divide  it  internally 
at  M  in  extreme  and  mean  ratio  so  that 

OA:  0M==  OM.AM. 

2.  With  A  as  center  and  OM  (the  longer  segment)  as  radius, 
draw  an  arc  cutting  the  given  circle  at  B. 

Statement.     AB  is  -^^  of  the  circle,  and  AB  is  the  side  of 
the  regular  inscribed  decagon. 

Proof.     1.  Draw  OB  and  BM. 

2.  In  A  OAB  and  A  ABM: 

Z.A=AA; 

OA:AB  =  AB:  AM. 

[Substituting  AB  for  OM  in  the  proportion  of  step  1,  Construction.] 

3.  .-.  A  OAB  ~  A  ABM  §  280 

4.  .'.  Z1  =  ZA0B.  Why? 

5.  A  OAB  is  isosceles,  and  hence  A  ABM  is  isosceles. 

6.  .-.  0M=  AB  =  BM,  or  A  0MB  is  isosceles.     Why  ? 

7.  .:ZA0B=Z2. 

8.  Z  4  =  Z  2  +  Z  AOB,  or  Z  4  =  2  •  Z  AOB.  Why? 

9.  .'.  Z3  =  2'ZAOB.  Why? 

10.  .:  ZAB0  =  2'ZA0B.  Why? 

11.  ZAOB  +  Z3  +  ZABO=1SO°.  Why? 


REGULAR  POLYGONS  233 

12.  .',5-^A0B  =  180°.  Why  ? 

13.  ..ZA0B=S6°. 

14.  Hence  AB  =  y^^  of  the  circle  and  AB  is  one  side  of  the 
regular  inscribed  decagon. 

Note. —  This  construction  is  attributed  to  Pappus. 

380.  Cor.  1.  Chords  joining  the  alternate  vertices  of  a  regu- 
lar inscribed  decagon,  starting  ivith  any  vertex,  form  a  regular 
inscribed  jyentagon. 

381.  Cor.  2.  Regular  inscribed  polygons  of  20,  Jf.0,  80,  etc., 
sides  can  be  constnicted  with  ruler  and  compass  alone.        Why  ? 

Note.  —  By  §§  879,  380,  and  381,  regular  inscribed  polygons  the  num- 
ber of  whose  sides  is  a  number  of  the  form  5  •  2"  can  be  constructed  with 
ruler  and  compass  alone. 

( What  is  6  .  2»  when  ?iisO?  1?  2?  3?  etc.) 

Ex.  31.  In  a  circle  having  a  3  in.  radius  inscribe  a  regular  decagon 
making  all  of  the  constructions, 

(Keep  the  resulting  figure  for  use  in  later  exercises.) 

Ez.  32.  Construct  the  adjoining  figure,  having  the 
points  A  2 1  in.  from  the  center  and  the  points  B  3  in. 
from  the  center. 

(From  Ex.  31,  obtain  the  arcs  which  are  ^^  of  the 
larger  circle.) 

Ex.  33.     Prove  that  the  diagonals  of  a  regular  pentagon  are  equal. 

[Construct  the  pentagon  in  a  circle  of  radius  3  in.,  using  the  arcs  ob- 
tained in  Ex.  31. J 

Ex.  34.  Prove  that  diagonal  AC  oi  regular  pentagon  ABODE  is 
parallel  to  side  BE.     (Circumscribe  a  O  about  the  pentagon.) 

Ex.  35.  If  ABODE  is  a  regular  inscribed  pentagon  in  circle  O,  prove 
that  a  diameter  perpendicular  to  side  DE  passes  through  B  and  is  also 
the  perpendicular-bisector  of  the  diagonal  AG. 

Ex.  36.     If  the  diagonals  AC  and  BE  of  a  regular  inscribed  pentagon 
ABODE  intersect  at  F,  prove  that  A  ABF is  isosceles. 
Prove  also  that  A  AEF  is  isosceles. 

Ex.  37.  Construct  by  ruler  and  compass  alone  an  angle  of  36° ;  also 
an  angle  of  18°. 

Note.  —  Supplementary  Exercises  32  to  37,  p.  301,  can  be  studied  now. 


234  PLANE    GEOMETRY  —  BOOK  V 

Proposition   IX.     Problem 

382.  Inscribe  a  regular  pentadecagon  (15-^ow)  in  a 
circle. 

Given  O  MN. 

Required  to  inscribe  in  O  MN  a  pentadecagon. 

Analysis.     1.   The  central  Z  of  a  pentadecagon  =  ^-—  =  24°. 

2.  But  24°  =  60° -36°.  ^^ 

3.  This  suggests  a  combination  of  the  constructions  of  §  374 
and  §  379. 

Construction.  1.  Draw  chord  AB,  a  side  of  a  regular  in- 
scribed hexagon,  and  chord  AC,  a  side  of  a  regular  inscribed 
decagon. 

2.  Draw  chord  BC. 

Statement.  BC  is  one  side  of  the  regular  inscribed  penta- 
decagon. 

Proof.     1.   BC  =  {\  —  yV)  or  tt  of  the  circle.  Const. 

383.  Cor.  Regular  polygons  of  30,  60,  etc.,  sides  can  he  in- 
scribed in  a  circle  by  ruler  and  compass  alone. 

Note.  —  Regular  polygons,  the  number  of  whose  sides  is  a  number  of 
the  form  15  •  2",  where  n  is  an  integer,  can  be  constructed  with  ruler  and 
compass  alone.     (§§382  and  383.) 

384.  Combining  the  results  of  §§  373,  376,  381,  and  383,  it 
can  be  said  that  regular  polygons  of  2",  3  •  2%  5  •  2",  and  15  •  2" 
sides  (n  =  an  integer)  can  be  inscribed  in  a  circle  by  ruler  and 
compass  alone. 

Eac.  38.     How  large  is  the  vertex  angle  of  a  regular  pentadecagon  ? 

Ex.  39.  What  regular  polygons  having  a  number  of  sides  less  than 
100  can  be  constructed  by  ruler  and  compass  alone  ? 


REGULAR   POLYGONS 


235 


Proposition  X.     Theorem 

385.  Regular  polygons  of  the  same  number  of  sides 
are  similar. 


Hypothesis.     ABODE  and  A'B'C'iyE^  are  regular  polygons 
of  5  sides. 

Conclusion.     ABODE  -  A'B'O'D'E', 
Proof.     1.   The  polygons  are  mutually  equiangular. 
[Since  each  Z  of  each  polygon  is  (^)°  or  108°.] 

2.  Since  AB  =  5(7  =  OD,  etc.,  and  A'B'  =  B'C  =  O'D',  etc. 

.,^^  =  ^=_^,etc.  Ax.  6,  §51 

A'B'      B'O'      CD'' 

3.  .-.  the  polygons  have  their  homologous  sides  proportional. 

4.  .-.  ABODE  -  A'B'O'D'E'.  Why  ? 

Ex.  40.     Construct  a  square  having  given  one  of  its  diagonals. 
Ex.  41.    A  square  is  inscribed  in  a  circle  of  radius  B.    Prove  : 

(a)  s,  =  EV2',  (c)  a4  =  |V2; 

(6)  p4  =  4:By/2;  (d)  *4  =  2  i?^. 

Ex.  42.  In  the  figure  for  §  369,  prove  that  the  apothem  of  the  in- 
scribed polygon  becomes,  when  extended,  the  apothem  of  the  circum- 
scribed polygon. 

Ex.  43.  In  the  figure  for  §  367,  prove  that  the  radius  drawn  to  any 
vertex  Y  is  the  perpendicular  bisector  of  the  side  AB  of  the  inscribed 
polygon. 

Ex.  44.  Prove  that  the  sides  of  a  regular  x)olygon  circumscribed  about 
a  circle  are  bisected  by  the  points  of  tangency. 


236 


PLANE    GEOMETRY  —  BOOK   V 


Proposition  XI.     Theorem 

386.  The  perimeters  of  two  regular  polygons  of  the 
same  number  of  sides  have  the  same  ratio  as  their 
radii,  or  as  their  apothems. 


F'    B 


Hypothesis.  P  and  P'  are  the  perimeters,  R  and  P'  are 
the  radii,  and  r  and  r'  are  the  apothems  respectively  of  the 
regular  polygons  AC  and  A'C  of  the  same  number  of  sides. 

Conclusion.  -^  =  ^  =  L, 

P'     M'     / 

Proof.  1.  Let  0  and  0'  be  the  centers  of  polygons  AC  and 
A^C  respectively. 

Draw  radii  OA,  OB,  O'A',  and  O'B',  and  apothems  OF  and 
O'F'. 

Polygon  AC  ^  polygon  A'C\ 

' '  P'      A'B'' 
A  OAB  -  A  O'A'B'. 


AB' 


6. 


AB  _R 
~  R" 
P 
P' 


and  also 


AB 
A'B' 


R' 


§385 
§297 
§295 

Why? 

Why? 


Ex.  45.  The  perimeters  of  regular  inscribed  polygons  of  6  and  12 
sides  respectively  inscribed  in  a  circle  of  diameter  2  are  approximately 
6  in.  and  6.21  in.  respectively.  What  are  the  perimeters  of  regular  in- 
scribed polygons  of  6  and  12  sides  respectively  in  a  circle  of  diameter  4  ? 
of  diameter  7  ?  of  diameter  10  ? 


REGULAR   POLYGONS  237 

387.  Cor.  The  areas  of  two  regular  polygons  of  the  same 
number  of  sides  have  the  same  ratio  as  the  squares  of  their  radii 
or  as  the  squares  of  their  apothems. 

1.   Let  K  and  K'  represent  the  areas  of  the  polygons  AC 

and  A'C  respectively.  

Then  4^=4^.  §344 


K 
K' 

AB" 
A'B'' 

K 
K' 

2.  j^  =  fi-  = :!_.  See  step  5,  §  386 


Eac.  46.  The  perimeters  of  regular  polygons  of  4  and  8  sides  respec- 
tively circumscribed  about  a  circle  of  diameter  2  in.  are  8  in.  and  6.63  in. 
respectively.  What  are  the  perimeters  of  regular  circumscribed  polygons 
of  4  and  8  sides  respectively  circumscribed  about  a  circle  of  diameter  6  ? 
of  diameter  5  ? 

Ex.  47.  The  area  of  a  regular  hexagon  inscribed  in  a  circle  of  radius 
3  in.  is  23.38  sq.  in.  What  is  the  area  of  a  regular  hexagon  inscribed  in  a 
circle  of  radius  6  in.  ?  of  one  in  a  circle  of  radius  1  in.  ? 

Ex.  48.  The  area  of  a  regular  octagon  circumscribed  about  a  circle 
of  radius  1  in.  is  1.656  sq.  in.  What  is  the  area  of  a  regular  octagon  cir- 
cumscribed about  a  circle  of  radius  2  in.  ?  of  radius  5  in.  ? 

Ex.  49.  Prove  that  the  square  inscribed  in  a  semi- 
circle is  equal  to  two  fifths  the  square  inscribed  in  the 
entire  circle. 

Sugrjestions,  —  Let  R  =  the  radius  of  the  circle.  Com- 
pute the  areas  of  the  two  squares. 

Ex.  50.  Prove  that  the  perimeter  of  any  regular  inscribed  polygon  is 
less  than  the  perimeter  of  the  regular  circumscribed  polygon  of  the  same 
number  of  sides. 

Ex.  51.  In  a  circle  of  2  in.  radius,  inscribe  a  square,  a  regular  octa- 
gon, and  also  a  regular  16-gon.    Prove  thatp4  <  p%  <pi6. 

Ex.  52.     About  a  circle  of  2  in.  radius  circumscribe  a  square,  a  regu- 
lar octagon,  and  a  regular  16-gon.     Prove  P4  >  Pg  >  Pie- 
Ex.  53.     (a)  Prove  that  p4,  pg,  Pie,  etc.,  are  each  less  than  P4. 
(6)  Prove  that  P4,  Pg,  Pie,  etc.,  are  each  greater  than  ^4. 

Suggestions.  —  1.  Compare  pi^  with  Pig.    See  Ex.  50. 
2.  Compare  Pie  with  P4. 


238 


PLANE    GEOMETRY  —  BOOK   V 


MENSURATION  OF  A  CIRCLE.     INFORMAL  TREATMENT 

388.  Length  of  a  Circle.  We  have  defined  the  length  of  a 
straight  line  segment  as  the  ratio  of  that  segment  to  the  unit  of 
linear  measure,  —  another  straight  line  segment.  Clearly  we 
cannot  define  the  length  of  a  circle,  in  that  manner,  because  we 
cannot  lay  off  the  linear  unit  of  measure  along  a  circle.  In 
defining  the  length  of  a  circle  therefore,  an  entirely  new  pro- 
cedure is  necessary.  The  treatment  which  follows,  while  in- 
formal, involves  nevertheless  the  ideas  which  underlie  the 
formal  treatment  of  this  same  topic  given  in  §  401  to  §  413 
inclusive. 

(a)  In  the  adjoining  circle  are  inscribed 
a  square  and  a  regular  octagon  ;  imagine 
that  the  regular  inscribed  polygons  of  16, 
32,  etc.,  sides  also  are  drawn. 

The  perimeters  of  these  polygons  have 
been  denoted  by  p^,  p^,  piQ,  etc.  (§  362). 

We  have  proved  (Ex.  51)  tliSit p8> p^; 
that  piQ  >  ps ;  that  ^32  >  PiQ ;  etc.    In  other 
words,  the  perimeters  of  the  regular  inscribed  polygons  increase 
as  the  number  of  sides  increases. 

(6)  About  the  adjoining  circle  there  are 
circumscribed  the  regular  polygons  of  4 
and  8  sides.  Imagine  that  those  of  16, 
32,  etc.,  sides  also  are  drawn. 

The  perimeters  of  these  polygons  have 
been  denoted  by  P4,  Pg?  Pm  etc.  (§  362). 

We  have  proved  (Ex.  52)  that  P,>  Ps; 
that  Pg  >  P16 ;  that  Pig  >  P32 ;  etc.  In  other  words,  the  perim- 
eters of  the  regular  circumscribed  polygons  decrease  as  the 
number  of  sides  increases. 

(c)  From  the  figure  it  is  evident  that  the  successive  inscribed 
polygons  come  closer  and  closer  to  the  circle  ;  likewise  that  the 
successive  circumscribed  polygons  come  closer  and  closer  to  the 
circle. 


MEASUREMENT   OF  THE  CIRCLE  239 

It  is  evident  also  that  the  length  of  the  circle  is  greater  than 
the  perimeter  of  any  inscribed  polygon  and  that  the  length  of  the 
circle  is  less  thay  the  perimeter  of  any  circumscribed  polygon. 

It  is  natural  therefore  to  regard  the  successive  perimeters  of 
the  regular  inscribed  polygons  and  also  of  the  regular  circum- 
scribed polygons  as  better  and  better  approximations  to  the 
length  of  the  circle. 

(d)  By  careful  computation  it  has  been  found  that  when  the 
diameter  of  a  circle  is  1 : 

p,    =2.82843.  P,   =4. 

^8   =3.06147.  Pg    =3.31371. 

j9i6  =  3.12145.  Pi6  =  3.18260. 

^32  =  3.13655.  P32  =  3.15172. 

p^  =  3.14033.  P64  =  3.14412. 

i)i28  =  3.14128.  Pi28  =  3.14222. 

jP26e  =  3.14151.  P256  =  3.14175. 

2)512  =  3.14157.  P512  =  3.14163. 

Apparently  when  the  diameter  of  a  circle  is  1,  the  length  of 
the  circle  is  approximately  3.1416.  If  we  let  C=the  length 
of  the  circle  and  d  =  the  length  of  the  diameter,  then  C  -i-d  — 
3.1416. 

(e)  By  Proposition  XI,  §  386,  the  perimeters  of  regular  poly- 
gons of  the  same  number  of  sides  have  the  same  ratio  as  their 
radii,  and  hence  as  their  diameters,  and  also  as  their  apothems. 

If  we  double  the  diameter  of  the  circle  considered  in 
part  (d),  then  we  shall  obtain  for  the  successive  perimeters  of 
the  inscribed  and  of  the  circumscribed  polygons  exactly  double 
the  lengths  given  in  part  (d).  Evidently  then  the  length  of  a 
circle  of  diameter  2  is  approximately  double  that  of  a  circle  of 
diameter  1 ;  that  is,  C  =  2  x  3.1416  =  6.2832.  Again,  C-^d  = 
3.1416. 

Similarly  the  length  of  a  circle  of  diameter  5  is  approxi- 
mately 5  X  3.1416,  or  15.7080.     Again  G^d  =  3.1416. 

389.  The  relation  derived  in  parts  (d)  and  (e)  of  §  388  is 
not  only  apparently  true  but  can  be  proved  to  be  true.     We 


240  PLANE    GEOMETRY  ~  BOOK  V 

shall  assume  it  for  the  present.  It  amounts  to  assuming  that 
the  length  of  a  circle  bears  to  the  length  of  its  diameter  a  con- 
stant ratio.     This  fact  is  proved  in  §  415. 

The  Greek  letter  tt  (pi)  is  used  to  denote  this  constant  ratio. 

That  is,  C^d=7r,  or  C=7rd. 

Two  useful  approximations  of  tt  are  3.1416  and  3^. 

The  length  of  a  circle  is  called  the  Circumference  of  the  circle. 

Note.  — The  determination  of  the  value  of  ir  and  of  what  sort  of  num- 
ber TT  is  has  been  one  of  the  most  famous  problems  of  mathematics. 

The  Egyptians  early  recognized  that  C  -^  d  i^  constant,  and  obtained 
for  this  ratio  a  value  which  corresponds  to  3.1605. 

The  Babylonians  and  Hebrews  were  content  with  the  much  less  accurate 
value,  TT  =  3.     (See  I  Kings,  vii.  23.) 

The  method  employed  in  this  text  was  introduced  by  Antiphon  (469- 
399  B.C.),  improved  by  Bryson  (a  contemporary,  probably),  and  finally 
carried  out  arithmetically  in  a  remarkable  manner  by  Archimedes  (287- 
212  B.C.)  in  a  pamphlet  on  the  mensuration  of  the  circle.  Antiphon 
suggested  the  use  of  inscribed  regular  polygons  of  4,  8,  etc.,  sides  as  a 
means  of  approximating  the  length  of  the  circle,  and  Bryson  suggested 
using  at  the  same  time  the  corresponding  circumscribed  regular  polygons. 
Archimedes  employed  inscribed  and  circumscribed  regular  polygons  hav- 
ing 3,  6,  .-.96  sides  in  his  computation,  and  showed  that  tt  >3^^  and 

<^. 

The  methods  employed  by  Archimedes  remained  for  a  long  time  the 
standard  procedure  in  efforts  to  compute  tt.  As  mathematical  skill  in- 
creased, formulse  for  ir  were  derived,  particularly  in  trigonometric  form, 
which  enabled  diligent  computers  to  obtain  the  value  to  more  and  more 
decimal  places. 

Vieta  (1540-1603)  was  the  first  to  derive  a  formula  for  tt  (not,  however, 
a  trigonometric  one).  He  gave  for  tt  the  value  3.141529653.  Others  car- 
ried out  the  computation  to  as  many  as  700  decimal  places. 

A  Holland  mathematician,  Huygens  (1629-1695),  at  the  age  of  twenty- 
five,  proved  some  theorems  which  made  it  possible  to  improve  greatly  on 
the  methods  of  Archimedes,  He  was  able  to  obtain  from  a  regular  hexagon 
as  accurate  a  value  for  tt  as  Archimedes  obtained  from  the  regular  96-gon. 

Mathematicians  were  particularly  interested  in  determining  what  kind 
of  number  tt  is.  In  1766-1767,  Lambert  proved  that  it  is  not  rational ; 
I.e.,  that  it  cannot  be  expressed  as  the  quotient  of  two  integers.  In  1882, 
through  methods  introduced  by  Hermite  in  1873,  Lindeman  proved  that 


MEASUREMENT   OF  THE   CIRCLE  241 

IT  is  a  transcendental  number ;  i.e. ,  that  it  cannot  be  the  root  of  an  ordinary 
algebraic  equation.  This  was  the  goal  toward  which  previous  efforts  had 
been  directed,  and  thus  completely  solved  a  problem  to  which  many  of  the 
great  mathematicians  had  given  some  attention. 

390.  Cor.  1.  The  circumference  of  a  circle  equals  2  ttt,  where 
r  equals  the  number  of  linear  units  in  the  radius. 

391.  Cor.  2.  The  circumferences  of  tivo  circles  have  the  same 
ratio  as  their  diameters  or  as  their  radii. 

Proof.  Let  r„  cZj,  and  Oj  be  the  radius,  diameter,  and  circum- 
ference of  one  circle ;  and  let  r^,  d^,  and  Cg  be  the  radius,  diam- 
eter, and  circumference  of  another  circle. 

2.   Then     C\  =  ttc^i  =  2  vTi,  and  Cj  =  7rc?2  =  2  Trrg. 

o  (7,     Trdi     2  7rr.         Ci      f7i      r, 

O.  .•.—!  = i  = 4,    or  — i=:— i  =  -^« 

Ca      TTfZz      ^Trrg  C2      di      r^ 

Note.  —  Remember  that  this  proof  is  based  on  an  informal  treatment. 
For  the  customary  formal  treatment  of  this  theorem,  read,  if  it  seems 
desirable,  §  414. 

Ex.  54.  Find  the  circumference  of  a  circle  whose  diameter  is  5  in. ; 
8  in. ;  10  in. 

Ex.  55.  How  long  is  the  piece  of  rubber  for  the  tire  of  a  buggy 
wheel  4  feet  in  diameter  ? 

Ex.  56.  If  the  diameter  of  a  circle  is  48  in.,  what  is  the  length  of  an 
arc  of  85°  ? 

Ex.  57.  How  long  must  the  diameter  of  a  circular  table  be  in  order 
to  seat  20  people,  allowing  30  in.  to  each  person  ? 

(Express  the  result  correct  to  the  nearest  inch.) 

Ex.  58.  A  fly  wheel  in  an  engine  room  has  a  diameter  of  10  feet. 
Through  how  many  feet  does  a  point  on  its  outer  rim  move  in  a  minute 
if  the  wheel  makes  100  revolutions  per  second  ? 

Ex.  59.  (a)  What  is  the  diameter  of  a  circular  race  track  whose 
length  along  its  inside  edge  is  one  mile  ? 

{h)  If  the  track  is  100  feet  wide,  determine  the  distance  around  it  in 
the  middle  of  the  track. 

Ex.  60.     Draw  any  circle.    Construct  the  circle  : 

(a)  Whose  circumference  is  3  times  that  of  the  given  circle.    See  §  391. 

(6)  Whose  circumference  is  \  that  of  the  given  circle. 


242  PLANE    GEOMETRY  —  BOOK  V 

392.  Area  of  a  Circle.  In  the  adjoining  circle  are  inscribed 
a  square  and  a  regular  octagon.  The  area  of  the  square  is  one 
half  the  product  of  its  apothem  and  its 
perimeter.     In  symbols  (§  362) : 

Similarly  the  area  of  the  regular  octa- 

And  the  area  of  the  regular  inscribed 
16-gon  would  be  ^ 

"'16  =  yPi6  X   Ct-16- 

It  is  evident  that  the  surface  within  each  successive  polygon 
is  more  nearly  equal  to  the  surface  within  the  circle.  On  the 
other  hand,  it  is  clear  that  each  successive  apothem  is  more 
nearly  equal  to  the  radius  and  that  the  length  of  the  polygon 
is  more  nearly  equal  to  the  length  of  the  circle.     (See  §  388.) 

It  is  reasonable  therefore  to  conclude  that  the  area  of  a  circle 
is  one  half  the  product  of  its  radius  and  its  circumference. 

Letting  K  represent  the  area  of  the  circle,  then 

393.  Cor.  1.     Since  (7  =  2  Trr,  then  7^  =  1  r  x  2  ttt-  =  ttt^. 

394.  Cor.  2.     Since  C  =  nd,  and  r  =  -,  then 

K=lxix7rd  =  ^7rd\ 

2      2  4 

395.  Cor.  3.  Tlie  areas  of  two  circles  have  the  same  ratio  as 
the  squares  of  their  radii  or  of  their  diameters. 

Letting  K^  and  K^  represent  the  areas  of  the  circles  whose 
diameters  are  d^  and  c?2j  ^^id  whose  radii  are  ri  and  rg  respec- 
tively,  then    ^,  ^  ^  ^  i^.^  ^^.  ^  ^  ^  ^  ^_ 
^2      Trrj^      \Trd^^        7^2       r^      d^ 

Note.  —  This  theorem  was  proved  by  Hippocrates  (450-400  b.c.  ).  Look  • 
up  his  history. 


MEASUREMENT  OF  THE  CIRCLE 


243 


396.  A  Sector  of  a  Circle  is  the  portion 
of  the  interior  of  a  circle  which  is  within 
a  given  central  angle.  The  central  angle 
is  called  the  angle  of  the  sector. 

397.  Cor.  4.  Tlie  area  of  a  sector  is  one 
half  the  i)Toduct  of  the  radius  and  the  length 
of  the  arc  intercepted  by  its  angle. 

Let  c  =  the  length  of  the  arc  and  k  =  the  area  of  the  sector 
of  a  circle  whose  area,  circumference,  and  radius  are  K,  C,  and 
r,  respectively. 


Prove  that 


k  =  ^r  X  c. 


Proof.    The  area  of  a  sector  has  the  same  ratio  to  the  area  of  the  circle 
that  the  length  of  its  arc  has  to  the  circumference  ;  that  is, 


^=^,  orJfc 


But,  since 
Substituting  in  (1), 


K 
K 


'"^  C 


(1) 


^-•f=i^- 


k  =  c  X  ^r,  or  ^r  x  c. 


398.  A  Segment  of  a  Circle  is  that  portion  of  the  interior  of 
a  circle  which  is  between  a  chord  of  the  circle 
and  its  subtended  arc ;  as  segment  AXB,  in- 
dicated by  the  shaded  part  of  the  adjoining 
figure. 

The  area  of  a  segment  AXB  may  be  deter- 
mined by  subtracting  the  area  of  A  AOB 
from  the  area  of  sector  OAXB. 


Ex.  61.    Find  the  circumference  and  area  of  a  circle  whose  diameter 
is  6  in.  ;  8  in.  ;  10  in. 

Ex.  62.     Find  the  radius  and  area  of  a  circle  whose  circumference  is 

20  TT  in.  ;  38  tt  in.  ;  15  tt  in. 

Ex.  63.     Find  the  radius  and  circumference  of  a  circle  whose  area  is 
64  T  sq.  in.  ;  81  tt  sq.  in.  ;  225 tt  sq.  in. ;  289 tt  sq.  in. 

Ex.  64.     Find  the  side  of  a  square  equivalent  to  a  circle  whose  diame- 
ter is  12  in. 


244  PLANE    GEOMETRY  —  BOOK  V 

Ex.  65.     The  diameters  of  two  circles  are  6  and  8  respectively. 

(a)  What  is  the  ratio  of  their  areas  ? 

(6)  What  is  the  ratio  of  their  circumferences  ? 

Ex.  66.  The  radii  of  three  circles  are  3,  4,  and  12,  respectively. 
What  is  the  radius  of  a  circle  equal  to  their  sum  ? 

Ex.  67.  rind  the  area  of  a  segment  having  for  its  chord  a  side  of  a 
regular  inscribed  hexagon,  if  the  radius  of  the  circle  is  10.  See  §  398 

Ex.  68-  If  the  radius  of  a  circle  is  4,  what  is  the  area  of  a  segment 
whose  arc  is  120°? 

Ex.  69.     Draw  any  circle. 

(a)  Construct  the  circle  whose  area  is  four  times  that  of  the  given 
circle.  §  395 

(6)  Construct  the  circle  whose  area  is  i  that  of  the  given  circle. 

Ex.  70.  Two  pulleys  in  a  machine  shop  are 
connected  by  a  belt.  One  has  a  radius  of  9  in. 
and  the  other  a  radius  of  1  in.  For  each  revolution 
of  the  large  pulley  how  many  revolutions  will  the 
small  pulley  make  ? 

Ex.  71.  What  is  the  area  of  the  ring  between  two  concentric  circles 
whose  radii  are  8  in.  and  10  in.  respectively  ? 

Ex.  72.  A  circular  grass  plot,  100  ft.  in  diameter,  is  surrounded  by 
a  walk  4  ft.  wide.     Find  the  area  of  the  walk. 

Ex.  73.  How  many  tulip  bulbs  will  be  required  for  a  circular  flower 
bed  6  feet  in  diameter,  allowing  16  sq.  in.  to  eacli  bulb  ? 

Ex.  74.  In  a  steam  engine  having  a  piston  20  in.  in  diameter,  the 
pressure  upon  the  piston  is  90  lb.  to  the  square  inch.  What  is  the  total 
pressure  upon  the  piston  ? 

Ex.  75.  A  woman  had  a  number  of  potted  plants  with  which  to 
plant  a  circular  flower  bed.  She  planned  to  make  the  bed  4  feet  in 
diameter  and  found  that  she  used  up  in  that  way  just  one  half  of  her 
plants.  Approximately  how  large  should  she  make  the  bed  to  use  up  all 
of  her  plants  ? 

Ex.  76.     Prove  that  the  area  of  the  ring  included  be-  a /^IIA^Xc 

tween  two  concentric  circles  is  equal  to  the  area  of  a  f  'yC  \  ^^\\ 

circle  whose  diameter  is  that  chord  of  the  outer  circle  j    i     n       j    j 

which  is  tangent  to  the  inner.  V  V  J  j 

(To  prove  area  of  ring  =  \  irAC'-^-)  \.^^^^ 


MEASUREMENT  OF  THE  CIRCLE 


245 


Ex.  77.  In  erecting  a  hot  air  furnace  for  dwellings,  certain  pipes  are 
installed  for  carrying  the  air  to  the  various  rooms  of  the  house,  and  one 
or  more  other  pipes  are  put  in  to  convey  cold  air  to  the  furnace.  The 
cross  section  area  of  tlie  cold  air  supply  pipes  must  equal  approximately 
tlie  sum  of  the  cross  section  areas  of  the  warm  air  pipes. 

A  house  is  to  have  four  warm  air  pipes  9  in.  in  diameter,  and  three  12  in. 
in  diameter.  One  cylindrical  cold  air  duct  is  to  be  installed.  How  large, 
approximately,  must  its  diameter  be  ? 

Ex.  78.  Prove  that  the  area  of  a  circle  is  equal  to  four  times  the  area 
of  the  circle  described  upon  its  radius  as  a  diameter. 

Ex.  79.  In  the  adjoining  semicircular  arch  con- 
structed about  center  0,  the  distance  AB  is  10  ft.  If 
the  arch  is  to  be  constructed  of  13  stones  of  equal  size, 
how  long  is  each  of  the  arcs  like  arc  DE  ? 


Ex.  80.  The  adjoining  figure  represents  a  segmen- 
tal arch.  The  method  of  construction  and  the  dimen- 
sions are  indicated  in  the  figure.  If  the  arch  is  made 
of  11  stones  of  equal  size,  what  is  the  length  of  the  arc 
Xr?     What  is  the  height  of  the  arch  ? 

Note.  —  Supplementary  Exercises  38  to  58,  p.  302, 
can  be  studied  now. 


SUPPLEMENTARY  TOPICS 

Four  groups  of  supplementary  topics  follow.  Each  is  inde- 
pendent of  the  others.  Teachers  should  feel  free  to  select  the 
group  or  groups  which  appear  to  meet  the  needs  of  the  class. 

Group  A.  —  Inscription  of  Circles  within  Regular  Polygons 
and  within  Circles. 

A  topic  of  considerable  interest  because  of  its  frequent  application  in 
artistic  design. 

Group  B. — Variables  and  Limits  together  with  the  Formal 
Treatment  of  the  Mensuration  of  the  Circle  and  of  the  Incom- 
mensurable Cases. 

The  treatment  of  this  topic  is  scientifically  correct,  but  is  nevertheless 
as  elementary  and  pedagogical  as  the  nature  of  the  subject  renders  possible. 

Group  C.  —  Symmetry  in  Plane  Figures. 

Group  D.  —  Maxima  and  Minima  of  Plane  Figures. 


246  PLANE    GEOMETRY  —  BOOK  V 

Group  A 

399.  Inscription  of  Circles  within  Regular  Polygons  and 
within  Circles  is  a  characteristic  feature  of  art  wIdcIow  and 
other  designs. 

Many  of  the  necessary  constructions  are  based  upon  the  fol- 
lowing illustrative  problem  or  may  be  discovered  by  means  of 
an  analysis  similar  to  that  employed  in  this  problem. 

Illustrative  Problem.  —  Inscribe  a  circle  in  a  given  sector 
of  a  circle. 

Analysis.     Let  O  XTZ  be  tangent  to  0 

radius  OB  at  Z,  to  OA  at  Y,  and  to  arc  /j\ 

AXB  at  X.     Let  CD  be  the  common  tan-  /  |  \ 

gent  to  arc  AXB  and  O  XYZ  at  X.  /H"~^ 

2.  Then  O  XFZ  is  inscribed  in  A  OOZ).  YJ       i      Az 

3.  Hence  the  center  P  of  Q  XYZ  lies  A  J:j        )\ 
on  the  bisectors  of  Z  COD  and  Z  OCD.  ^,/<<      I  ^JJ^ 

The  radius  is  the  distance  from  P  to  X.    O  X  ^ 

The  construction  is  evident  at  once. 

Rs.  81.  Construct  a  circle  with  radius  2  in. ;  and  within  it  construct 
a  sector  whose  angle  is  90°. 

(a)  Within  this  sector  inscribe  a  circle. 

(?))  Compare  the  area  of  this  circle  with  the  area  of  the  sector  itself 
when  the  radius  of  the  given  circle  is  r  instead  of  2. 

Ex.  82.  Construct  six  equal  circles  within  a  circle  of 
radius  2  in.,  each  tangent  internally  to  the  given  circle 
and  tangent  externally  to  two  of  the  inner  circles. 

Ex.  83.   (a)  If  the  radius  of  the  given  circle  in  Ex.  82 
is  r,  what  is  the  radius  of  the  inscribed  circles  ? 

(&)  Compare  the  circumference  of  one  of  the  inscribed  circles  with  the 
circumference  of  the  given  circle. 

(c)  Compare  the  total  area  of  the  inscribed  circles  with  the  area  of  the 
given  circle. 

Ex.  84.  Construct  a  circle  which  will  be  tangent  to  each  of  the  con- 
structed circles  in  Ex.  82. 

How  does  the  radius  of  this  circle  compare  with  the  radius  of  the  six 
inscribed  circles  ? 


INSCRIBED  CIRCLES 


247 


Ex.  85.  Construct  six  equal  circles  tangent  externally 
to  a  circle  of  radius  ^  in.  such  that  each  circle  is  also 
tangent  externally  to  two  of  the  constructed  circles. 

Ex.  86.  If  the  radius  of  the  given  circle  in  Ex.  85 
is  r: 

(a)  What  is  the  radius  of  the  escribed  circles  ? 

(6)  Compare  the  circumference  of  the  given  circle  with  the  circumfer- 
ence of  one  of  the  escribed  circles. 

(c)  Compare  the  area  of  the  given  circle  with  the  area  of  one  of  the 
escribed  circles. 

Ex.  87.  In  an  equilateral  triangle  inscribe  three  equal 
circles,  each  tangent  to  two  sides  of  the  triangle  and  tan- 
gent externally  to  the  other  two  circles. 

Ex.  88.  In  a  regular  hexagon  inscribe  six  equal  cir- 
cles, each  tangent  to  two  sides  of  the  hexagon  and  tangent 
externally  to  two  of  the  circles. 

Ex.  89.  In  a  regular  octagon  inscribe  eight  equal  circles,  each  tangent 
to  two  sides  of  the  octagon  and  also  tangent  externally  to  two  of  the 
circles. 

Ex.  90.  In  a  regular  hexagon  inscribe  six  equal  circles,  each  tangent 
to  one  side  of  the  hexagon  and  tangent  externally  to  two  of  the  circles. 

Ex.  91.  Inscribe  in  a  regular  hexagon  three  equal  circles 
each  tangent  to  two  sides  of  the  hexagon  and  tangent  exter- 
nally to  two  circles. 

Ex.  92.  In  a  regular  octagon  inscribe  four  equal  circles 
each  tangent  to  two  sides  of  the  octagon  and  also  tangent  ex- 
ternally to  two  circles. 

Ex.  93.  The  adjoining  design  appears  in  a  floor 
pattern  in  a  corridor  of  the  new  Congressional  Li- 
brary. Construct  such  a  figure,  making  a  5-in.  square, 
the  radius  of  the  inner  O  1.5  in.,  and  the  radius  of  the 
concentric  O  1.76  in. 

Ex.  94.   The  adjoining  curve  is  a  trefoil. 

(a)  Construct  such  a  figure  based  upon  an  equilateral 
triangle  whose  side  is  2  in.  long. 

(b)  What  is  the  length  of  the  trefoil  if  the  side  of  the 
equilateral  triangle  is  s  inches  ? 

(c)  What  is  the  area  within  the  trefoil  if  the  length 
of  the  side  of  the  equilateral  triangle  is  s  inches  ? 

Note.  —  Supplementary  Exercises  59  to  63,  p.  304,  can  be  studied  now. 


248  PLANE    GEOMETRY  —  BOOK   V 


Group  B.     Mensuration  of  a  Circle 

400.  The  formal  treatment  of  the  mensuration  of  a  circle 
involves  the  use  of  certain  ideas  which  are  fundamental  in 
mathematics. 

401.  Variable,  Constant,  and  Limit. 

Example  1.  —  Consider  the  numbers  1,  -,  -,  -,  ... 

'  2'  4'  8' 

Each  number  is  one  half  the  preceding;  while  each  number 

is  greater  than  zero,  the  numbers  ultimately  become  very  small. 

Imagine  a  literal  number  x  which  has  these  values  successively. 

Then  ultimately  x  —  0  becomes  less  than  an}^  small  positive 

number,  and  thereafter  remains  less  than  that  number.     Thus, 

ultimately,   a;  — 0   becomes   and   remains  less   than  ,    or 

1       •^'  _  1000' 

■ ,  or  any  other  small  positive  number. 

1,000,000  -^  ^ 

This  is  clear,  since  x  takes  successively  the  values 

.1111111         1         1  1 

J     _    _    . etc 

'    2'  4'  8'    16'  32'  64'   128'  256'  512'   1024' 


Example  2.  —  Consider  the  numbers  1,  1^,  1|,  l-J,  1||,  ... 
Although  each  number  is  less  than  2,  the  numbers  are  constantly 
increasing. 

Imagine  a  literal  number  x  which  has  these  values  succes- 
sively. Then  ultimately  2  —  x  becomes  and  remains  less  than 
any  small  positive  number  ;  thus  2  —  x  becomes  and  remains 
less  tinan  ■yq  o  o7' 

(Write  down  enough  of  the  successive  values  of  x  to  make 
certain  of  this  last  statement.) 

A  Variable  is  a  number  which  assumes  different  values 
during  a  particular  discussion.  Thus,  in  Example  1,  a;  is  a 
variable  ;  it  is  a  decreasing  variable.  In  Example  2,  x  is  an  in- 
creasing variable. 

Note.  —  Variables  do  not  either  always  increase  or  always  decrease. 
Thus,  the  variable  which  takes  the  values  1,  —  |,  +  \^  —  |,  etc.,  al- 
ternately decreases  and  increases. 


VARIABLES  AND  LIMITS  249 

A  Constant  is  a  number  which  has  a  fixed  value  tliroughout  a 
particular  discussion.  Thus,  in  Example  2,  2  is  constant ;  in 
Example  1,  0  is  constant. 

A  Limit  of  a  Variable  is  a  constant  such  that  the  numerical 
value  of  the  difference  between  the  constant  and  the  variable 
becomes  and  remains  less  than  any  small  positive  number. 

We  say  that  a  variable  approaches  its  limit. 

Not  every  variable  has  a  limit. 

402.  Axiom  of  Limits.  If  an  increasing  variable  is  always 
less  than  some  constant^  then  it.  approaches  a  limit  which  is  less 
than  or  equal  to  that  constant.  Jf  a  decreasing  variable  is  always 
greater  than  some  constant,  then  it  approaches  a  limit  ivhich  is 
greater  than  or  equal  to  that  constant. 

We  may  represent  the  foregoing  definitions  and  axiom  geo- 
metrically as  follows : 

X 

A  z  (J  B 

Let  the  distance  from  A  to  X  as  X  moves  toward  B  represent 
a  variable  x.  Let  it  be  agreed  that  X  never  passes  beyond  B. 
Then  ^X  must  approach  a  limit  such  as  AC  which  is  less  than 
or  equal  to  AB.  (In  the  figure,  AC  is  made  less  than  AB.) 
This  means  that  ultimately  point  X  comes  and  remains  as  close 
to  C  as  we  please,  possibly  even  coinciding  with  C. 

403.  Two  Limits  Theorems. 

(a)  If  a  variable  x  approaches  a  finite  limit  I,  then  ex,  where  c 
is  a  constant,  approaches  the  limit  d. 

For  cl  —  cx  =  c(l  —  x).  As  X  approaches  the  limit  Z,  the  numerical 
value  of  Z  —  X  becomes  and  remains  less  than  any  small  positive  number. 
Hence  the  numerical  value  of  cl  —  ex  becomes  and  remains  less  than  any 
small  positive  number.    Therefore  ex  approaches  the  limit  cl  by  definition. 

(b)  If  two  variables  are  constantly  equal  and  each  approaches  a 
finite  limit,  then  their  limits  are  equal. 


250  PLANE    GEOMETRY  —  BOOK  V 

Let  X  approach  the  limit  I  and  y  approach  the  limit  m.  If  x 
always  equals  y,  then  /  must  equal  m. 

A ^ ? ^ L 

^- y P ^ 

Let  the  distance  AX  represent  the  variable  x,  approaching 
AL.  Let  the  distance  B  Y  represent  the  variable  y,  approach- 
ing the  limit  BM. 

Point  X  ultimately  must  come  and  remain  close  to  the  point 
L  ;  point  Y  ultimately  must  come  and  remain  close  to  the  point 
M.  If  AL  were  greater  than  BM,  then  AX  would  ultimately 
become  greater  than  BY.  But  this  is  impossible,  for  the  vari- 
able X  must  always  equal  the  variable  y.  Similarly  if  AL  were 
less  than  BM.     Hence  the  limit  I  equals  the  limit  m. 

404.  Sequences  of  Regular  Polygons.     Let  AB  be  a  side  of 

a  regular  inscribed  polygon  in  a  circle  of  radius         ^ -^ 

r;   let  AC  be  a  side   of   the  regular   polygon  /^  \ 

having   double   the   number   of   sides    of    the  (?) 

first;  let  AD  be  a  side  of  the  regular  polygon  V        /j  J 

having   double   the   number   of   sides   of    the  ^X>^i  ^^^ 
second;  etc.  0 

Such  regular  inscribed  polygons,  the  number  of  whose  sides 
is  successively  doubled,  will  be  called  a  sequence  of  regular 
inscribed  polygons. 

Similarly,  we  shall  have  occasion  to  speak  of  sequences  of 
regular  circumscribed  polygons. 

405.  In  a  sequence  of  regular  inscribed  polygons,  the  length  of 
the  side  of  the  polygon  is  a  decreasing  variable  which  approaches 
zero  as  limit,  as  the  number  of  sides  increases  indefinitely. 

For  3^rc  AC  =  I  arc  AB  (see  Fig.  §  404)  ;  arc  AD  =  J  arc  AB  ;  etc. 

Hence  the  arcs  decrease  indefinitely,  approaching  zero  as  limit.  (See 
Ex.  1,  §401.) 

The  chords  are  less  than  the  corresponding  arcs.  Hence  the  chordg 
decrease  indefinitely,  approaching  zero  as  limit. 


VARIABLES  AND  LIMITS 


251 


Evidently,  also  in  a  sequence  of  regular  circumscribed  poly- 
gons the  length  of  the  side  is  a  decreasing  variable  which 
approaches  zero  as  limit  as  the  number  of  sides  increases 
indefinitely. 

406.  In  a  sequence  of  regular  inscribed  polygons,  the  length 
of  the  apothem  is  an  increasing  variable  which  approaches  the 
radius  of  the  circle  as  limit. 

If  ^5  is  a  side  of  any  regular  inscribed  polygon  and 
00  is  the  apothem  of  the  polygon,  then 

AO-  0C<  AC,  or  AO-  00 <^AB.     (§160.) 

As  the  number  of  sides  increases  indefinitely,  AB 
decreases  indefinitely  in  numerical  value.  Hence 
AO  —  00  must  also  decrease  indefinitely  in  numerical 
value.     Therefore,  00  must  approach  AO  as  limit  by  definition. 


LENGTH  OF  A  CIRCLE 

407.  Consider  the  sequences  of  regular  inscribed  and  circum- 
scribed polygons  having  4}  S,  16,  etc.,  sides  in  a  circle  of  radius  r. 

Let  p  denote  the  variable  perimeter  of  the  inscribed  poly- 
gon, assuming  the  values  p^,  pg,  pi^,  etc. ;  let  P  denote  the 
variable  perimeter  of  the  circumscribed  polygon,  assuming  the 
values  of  P^,  Pg,  P^,,  etc. 


i 

'  .    ^ 

>- 

-r^v 

r 

^ 

X 

•0       z 

V 

wy 

K 


408.    The  perimeter  p  approaches  a  limit  as  the  jiumber  of 
sides  increases  indefinitely. 

Proof.     1.  Pi  <ps  <pi6,  etc.  See  Ex.  51 

2.  1>4,  P8,  Pi6,  etc.  are  all  less  than  P4.  See  Ex.  63,  (a) 

3.  .-.p  approaches  a  limit  as  the  number  of  sides  increases  indefinitely. 
Call  this  limit  ^4.  §  402 


252  PLANE    GEOMETRY  —  BOOK   V 

409.  The  j^erimeter  P  (see  §  407)  approaches  a  limit  as  the 
number  of  sides  increases  indefinitely. 

Proof.     1.    P4  >  Pg  >  P16,  etc.  See  Ex.  52 

2.  P4,  Pg,  P16,  etc.  are  each  greater  than  jt)4.  See  Ex.  53,  (6) 

3.  .'.  P  approaches  a  limit  as  the  number  of  sides  increases  indefi- 
nitely.    Call  this  limit  L^.  §  402 

410.  The  perimeters  p  and  P  of  the  sequences  of  regular  in- 
scribed and  circumscribed  polygons  described  in  §  407  approach 
one  and  the  same  limit ;  that  is,  the  limit  l^  equals  the  limit  L^. 

Proof.     1.    Let  AB  =  a  side  of  one  of  the  inscribed  polygons,  and  AD 
and  DB  halves  of  two  consecutive  sides  of  the  circum- 
scribed polygon  having  the  same  number  of  sides.     OD 
and  OA  are  the  radii  of  these  polygons,    p  and  P  de- 
note their  perimeters. 

2.  The  two  polygons  are  similar. 

3.  .-.  P:p=  OD:  OA. 

4.  .-.  (P-p)  :p=(OD-  OA):  OA. 

5.  ,',P-p  =  J^(^OD-  OA) .  By  Algebra 

OA 

6.  .'.P-p<^x  AD. 

r 

(Since  J9  <  P4  ;   r  =  OA;  smd  OD  -  0A<  AD.) 

7.  Successively  double  the  number  of  sides,  letting  the  inscribed  and 
the  circumscribed  polygons  always  have  the  same  number  of  sides.  The 
length  of  each  side  of  the  polygons  will  decrease,  approaching  the  Umit 
zero  ;  in  particular,  AD  will  approach  0  as  limit. 

8.  .'.  -i  X  AD  will  approach  0  as  limit,  since  -^  is  constant. 

^  ^  §  403,  (a) 

9.  P  —  p  will  approach  0  as  limit.  Def.,  §  401 

10.  .-.  X4  =  Z4. 

For  suppose  that  X4  >  ^4,  and  that  ^4  —  ^4  =  m,  a  number  >  0.  Ulti- 
mately,' P  differs  but  little  from  X4  and  p  but  little  from  h  ;  hence  P  —  p 
ultimately  differs  but  little  from  m.     But  P  —  p  approaches  the  limit  0. 

Similarly  if  ^4  <  ^4. 

11.  Let  C  represent  the  common  value  of  L\  and  l^. 

Then,  as  the  number  of  sides  increases  indefinitely,  the  perimeters  p 
and  P  of  the  regular  inscribed  and  circumscribed  polygons  respectively 
approach  the  limit  C. 


VARIABLES  AND  LIMITS  253 

411.  If  now,  instead  of  starting  with  the  sequences  of  regu- 
lar polygons  having  4,  8,  16,  etc.  sides,  we  start  with  any  other 
sequences  of  regular  inscribed  polygons  and  circumscribed 
polygons,  such  as  those  having  3,  6,  12,  etc.  sides,  it  can  be 
proved  that  the  perimeters  p  and  P  again  approach  this  same 
limit  C  obtained  in  §  410. 

This  fact  justities  the  following  definition. 

412.  The  Length  of  a  Circle  is  the  limit  of  the  perimeter  of 
any  regular  inscribed  polygon  as  the  number  of  sides  is  in- 
definitely increased. 

Remember  that  the  length  of  a  circle  is  called  the  circum- 
ference of  the  circle. 

413.  The  perimeter  of  any  regular  circumscribed  polygon  ap- 
proaches the  circumference  of  the  circle  as  limit  if  the  number  of 
sides  is  indefinitely  increased.  §§  410,  411,  and  412 

Ex.  95.  What  is  a  constant  ? 
Ex.  96.  What  is  a  variable  ? 
Ex.  97.    What  is  the  limit  of  a  variable  ? 

Ex.  98.  If  a  "sequence"  of  regular  inscribed  polygons  be  formed 
(§  404)  in  a  circle  : 

(a)  What  magnitude  is  constant  ? 

(h)  What  magnitudes  are  decreasing  variables,  and  what  are  their 
limits  ? 

(c)  What  magnitudes  are  increasing  variables,  and  what  are  their 
limits  ? 

Ex.  99.  What  limit  is  approached  by  the  variable  which  assumes  the 
values  given  in  the  note  at  the  bottom  of  page  248  ? 

Ex.  100.     Suppose  a  variable  assumes  the  values  1,  —  1,  1,  —  1,  .... 

Does  it  approach  a  limit  ? 

Ex.  101.     What  is  the  limit  of  —  as  n  assumes  the  values  1,  2,  3, 

3n  '     '     ' 

4...? 


254  PLANE    GEOMETRY  —  BOOK   V 


Proposition  XII.     Theorem 

414.    Tlie  circumferences  of  two  circles  have  the  same  ratio  as 
their  radii  or  their  diameters. 


Hypothesis.  Ci  and  64  are  the  circumferences  of  two  circles 
whose  radii  are  r^  and  r^  and  whose  diameters  are  d^  and  c?2 
respectively. 

Conclusion.  O,  ^  n  ^  d^ 

Proof.  1.  Inscribe  in  the  circles  regular  polygons  having 
the  same  number  of  sides.  Let  p^  and  p2  be  the  perimeters 
of  the  polygons  inscribed  in  the  circles  whose  radii  are  7\  and 
rg  respectively. 

2.  The  polygons  are  similar.  §  385 

3.  Pi  =  h.  §386 

P2     ^2 

4.  .-.  p^  X  r2  =  P2  X  n.  Why  ? 

5.  Let  the  number  of  sides  of  each  polygon  be  successively 
doubled,  the  two  polygons  continuing  to  have  the  same  num- 
ber of  sides. 

6.  Then  pi  x  7*2  will  approach  the  limit  0^  X  ?*2 

and  P2  X  Ti  will  approach  the  limit  C^  X  r^. 
(§411,  §403,  (a)) 

7.  .-.  Ci  X  ^2  =  (72  X  ?'i.  §  403,  (6) 

8.  •••^=--  §252 

62     ^2 

9.  .-.       ^l^^Ti^di^  ^^^^ 

O2         ^  ^2         ^2 


VARIABLES  AND  LIMITS 


255 


415.   Cor. 


Since  ^^  =  ^,then  ^=^^ 


di     d^ 

That  is,  the  ratio  of  the  circumference  of  a  circle  to  the  length 
of  the  diameter  of  the  circle  is  constant  for  all  circles. 
Note.  —This  proves  the  fact  assumed  in  §  389. 

We  recall  that  the  constant  value  —  is  denoted  by  tt. 

d 

416.  Area  of  a  Circle.  The  formal  treatment  of  this  topic 
is  exactly  like  that  for  the  length  of  a  circle. 

Consider  the  sequences  of  regular  inscribed  and  circum- 
scribed polygons  in  a  circle  of  radius  r,  having  4,  8,  16,  etc. 
sides.     (§  407.) 

A\ ^    -     ^ 

E 
\B 

L 

Let  k  denote  the  variable  area  of  the  inscribed  polygon,  and 
K  the  variable  area  of  the  circumscribed  polygon  as  the  num- 
ber of  sides  increases. 

The  following  theorems  can  then  be  proved : 

(a)  The  area  k  approaches  a  limit  as  the  number  of  sides  in- 
creases indefinitely.     Call  this  limit  i^. 

(b)  The  area  K  approaches  a  limit  as  the  number  of  sides 
increases  indefinitely.     Call  this  limit  I^. 

(c)  The  limit  ii  =  the  limit  I^;  that  is,  the  areas  of  the  regu- 
lar inscribed  and  circumscribed  polygons  approach  the  same 
limit  as  the  number  of  sides  increases  indefinitely.  Call  this 
area  S. 

(d)  If  any  other  sequence  of  regular  inscribed  or  circum- 
scribed polygons  of  the  same  circle  be  formed,  the  areas  of  the 
inscribed  and  of  the  circumscribed  polygons  approach  the 
limit  S  obtained  in  the  theorem  (c). 


256  PLANE    GEOMETRY  —  BOOK  V 

417.  The  Area  of  a  Circle  is  the  limit  of  the  area  of  a 
regular  inscribed  polygon  as  the  number  of  sides  increases 
indefinitely. 

418.  Tlie  area  of  any  regular  circumscribed  polygon  ap- 
proaches the  area  of  the  circle  as  limit  as  the  number  of  sides  is 
increased  indefinitely.  This  follows  at  once  from  theorem  (d) 
of  §  416. 

Proposition  XIII.     Theorem 

419.  The  area  of  a  circle  is  the  product  of  one  half  its  radius 
and  its  circumference. 


Hypothesis,  r  is  the  radius,  C  is  the  circumference,  and  S 
is  the  area  of  the  circle. 

Conclusion.  zS  =  |  r  x  C. 

Proof.     1.   Circumscribe  about  the  circle  any  regular  polygon. 
Let  P  denote  its  perimeter  and  ^its  area. 

2.  Then,  since  its  apothem  is  r, 

/f=irxP.  Why? 

3.  Let  the  number  of  sides  of  the  circumscribed  polygon  be 
indefinitely  increased. 

Then  /^will  approach  the  limit  8\  §  418 

P  will  approach  the  limit  (7;  §  411 

i  rP  will  approach  the  limit  |  rO.         §  403,  (a) 

4.  .:S  =  ^r-C.  §403,(5) 

420.  The  Value  of  tt.  In  §  389,  the  approximate  value 
3.1416  for  TT  was  given.  This  value  was  derived  from  a  table 
of  values  of  perimeters  of  regular  inscribed  and  circumscribed 
polygons.  The  following  proposition  provides  a  means  of 
computing  such  tables  of  perimeters. 


LIMITS  AND  VARIABLES  257 

Proposition  XIV.     Problem 

421.  Given  p^  and  P„,  the  perimeters  of  the  regular  inscribed 
and  of  the  regular  circumscribed  polygons  having  n  sides ;  find 
;)2„  and  Po^j  ^^^^  perimeters  of  the  regular  inscribed  and  the 
regular  circumscribed  polygons  having  double  the  number  of 
sides. 

a!         M  F       V p' 


A. 


yB 


O 
(a)  To  find  P^^.  . 
Solution.     1.   Let  ^B  be  one  side  of  the  regular  inscribed 
polygon  having  n  sides  and  F  the.  mid-point  of  arc  AB, 

2.  Draw  the  tangent  to  the  circle  at  F,  meeting  OA  and  OB 
extended  at  A'  and  B^  respectively. 

Then  A'B^  is  one  side  of  the  regular  circumscribed  polygon 
having  n  sides.     Also  A'F=  ^  A'B',  and  hence  2nA'F=  P„. 

3.  Let  tangents  to  the  circle  at  A  and  B  meet  A'B'  at  3f 
and  ^respectively.  Then  MNis  one  side  of  the  regular  cir- 
cumscribed polygon  having  2 n  sides;  MF=^MN2ind  hence 


§270 


nMF=P^,. 

4. 

A'M     OA' 
MF      OF 

(Since  Oilf  bisects  Z  ^' OF. ) 

5. 

But    ^-^■ 

6. 

.    P,_A'M 

rr 

P,+p,     A'M+MF 

P.,               MF 

8. 

P..  +  P.     A'F 

2K          MF 

§386 
Why? 
Why? 
Why? 


258  PLANE    GEOMETRY  —  BOOK   V 

A'  M  F        N  B' 


A'\ 


Pn+Pn_^nA'F_2P^ 


or 


Pn 


4:nMF 


I  Multiplying  num.  and  denom.  of  ^^  by  4  w."| 


10. 

.•.i^2„(P.+K)=2PA. 

11. 

1. 

2. 

(b)  To  find  p2,.. 
A  ABF  and  A  AFM  are  isosceles  triangles 
ZABF:=ZAFM. 

3. 

.-.  A  ABF ^  A  AFM. 

4. 

AF:AB  =  MF:AF. 

5. 

.'.  AF' =  AB  X  MF. 

6. 

But  AF-  ^'- ;  AB-^'^;  and  MF  =  ^^^ . 
2n                 n                        An 

. 

[Steps  2  and  3,  part  (a)] 

P2^V_ 
jP2„ 


4/^2 


P_n 

n 
4tn^ 


•••  P2„ = yp„P2, 


Why? 

Algebra 


Why? 
Why? 
Why? 
Why? 
Why? 


Ax.  2,  §  51 

Algebra 
Algebra 


Note.  —  The    formulae  P2„  = ^^  and  ^2n  =  y/pnPin  are  quite  re- 

Pn  +  Pn 

markable.  If  the  perimeters  of  the  regular  inscribed  and  regular  circum- 
scribed polygons  of  say  4  sides  are  known,  then  the  perimeters  of  the 
regular  circumscribed  and  regular  inscribed  polygons  of  8  sides  can  be 
computed  by  mere  substitution  ;  then  those  of  16  sides  ;  and  so  on. 


VARIABLES  AND  LIMITS 


259 


Proposition  XV.     Problem 
422.    Compute  an  approximate  value  o/tt. 

Solution.  1.  If  the  diameter  of  a  circle  is  1,  the  side  of  an 
inscribed  square  is  ^  V2,  and  hence  the  perimeter  of  the  square 
is  2  V2,  or  p4  =  2.82843. 

2.   The  side  of  a  circumscribed  square  is  1,  and  P^  =  4. 

q                                                                       p_   2P4    Xi?4. 
tJ.  /-g—    -^- 

„      2  X  4  X  2.82843 


§421 


Hence 

4. 

Hence 


4  +  2.82843 


3.31371. 


P%  =  Vi)4  X  Pg- 


§421 


Ps  =  V2.82843  X  3.31371  =  3.06147. 


5.-  Similarly   P^^ 


2P,xps  _  2  X  3.31371  x  3.06147 


Pg  +  ^8  3.31371  +  3.06147 

=  3.18260. 
And    p,8  =  Vj98  X  P16  =  V3.06147  x  3.18260  =  3.12145. 

6.   In  this  manner,  we  compute  the  following  table  ; 


No.  OP 

Perimeter  of 

Perimeter  of 

Sides 

Reg.  Giro.  Polygon 

Reg.  Inso.  Polygon 

4 

4. 

2.82843 

8 

3.31371 

3.06147 

16 

3.18260 

3.12145 

82 

,  3.15172 

3.13655 

64 

3.14412 

3.14033 

128 

3.14222 

3.14128 

256 

3.14176 

3.14151 

512  . 

3.14163 

3.14157 

7.  The  last  results  show  that  the  circumference  of  the 
circle  whose  diameter  is  1  >  3.14157  and  <  3.14163. 

Hence  an  approximate  value  of  tt  is  3.1416,  correct  to  the 
fourth  decimal  place. 


260  PLANE    GEOMETRY  —  BOOK  V 

THE  INCOMMENSURABLE   CASES* 

Proposition  XVI.     Theorem 

423.  In  the  same  circle  or  in  equal  circles^  central  angles  have 
the  same  ratio  as  their  intercepted  arcs.  (When  the  angles  are 
incomm  ensurable.) 


Hypothesis.  In  O  ABC,  Z  AOB  and  Z  BOC  are  two  in- 
commensurable central  angles  intercepting  the  arcs  AB  and 
BC  respectively. 

Z  BOC     arc  BC 


Conclusion. 


ZAOB     arc  ^5 


Proof.  1.  Divide  Z  AOB  into  two  equal  parts  and  let  one 
of  these  be  applied  as  unit  of  measure  to  Z  BOC. 

2.  Since  Z  AOB  and  Z  BOC  are  incommensurable,  a  certain 
number  of  angles  equal  to  J-  Z  AOB  will  equal  Z  BOX^,  leav- 
ing a  remainder  Z  X^OC  which  is  less  than  the  unit  of 
measure. 

3.  Z  AOB  and  Z  BOX^  are  commensurable. 

^  Z  BOX^  ^  arc  BX, 

Z  AOB      arc  AB 

*Note. — There  are  three  incommensurable  cases.  (§423-§425.) 
These  propositions  complete  the  proofs  of  the  theorems  given  in  §  213, 
§  261,  and  §  327  respectively.  If  it  is  desired  to  read  §  423  when  studying 
§  213,  then  it  will  be  necessary  to  read  also  §  401  to  §  403  inclusive,  which 
give  an  introduction  to  the  theory  of  limits. 


VARIABLES  AND  LIMITS  261 

5.  Take  now  as  unit  of  measure  \  Z  AOB.  This  measure 
will  be  contained  an  integral  number  of  times  in  Z  AOB  and 
also  in  /.  BOX^^  further,  the  unit  of  measure  may  be  con- 
tained once  in  Z  XiOC,  leaving  a  remainder  Z  X^OC  which  is 
less  than  the  new  unit  of  measure. 

Again  Z^OX,  ^arc^X,. 

6.  Continue  in  this  manner  to  decrease  indefinitely  the  unit 
of  measure.  The  remainder  Z  XOC,  being  always  less  than 
the  unit  of  measure,  will  approach  the  limit  0. 

Using  the  symbol  =  to  express  "approaches  the  limit,'' 

Z  BOX  =  Z  BOC,  and  hence  yTS^ZJ^'  ^  ^^^'  ^^^ 

arc  BX  =  arc  BC,  and  hence  ^I^_BX^^c^C      ^  ^^^   ,  . 

arc^S     arc^B  '  ^  ^ 

7      ZBOX       .   SLTcBX  .  ,,  ,  .  ,  , 

'•     -^ — 7w^  ^-^d -—   are  variables  which   are   always 

ZAOB  arc  AB  ^ 

equal. 

o  .   Z  BOC     are  BC  ,,  .^o  /i.x 


262  PLANE    GEOMETRY  —  BOOK  V 

Proposition  XVII.     Theorem 

424.  A  parallel  to  one  side  of  a  triangle  divides  the  other  two 
sides  proportionally,  when  the  segments  of  one  side  are  incom- 
mensurable. 

A 


Hypothesis.  In  A  ABC,  segments  AD  and  BD  are  incom- 
mensurable ;  DE  II  BC,  meeting  AC  at  E. 

Conclusion.  ^^9^. 

AD     AE 

Proof.  1.  Divide  AD  into  any  number  of  equal  parts  (say- 
two),  and  apply  one  of  these  parts  to  BD  as  unit  of  measure. 

2.  Since  AD  and  BD  are  incommensurable,  a  certain  num- 
ber of  segments  equal  to  the  unit  of  measure  will  extend  from 
D  to  Xj,  leaving  a  remainder  XiB  which  is  less  than  the  unit 
of  measure. 

3.  Draw  X^Y^  II  BC,  meeting  AC  at   Y^. 

Then  :5^  =  ^S.  §261 

AD       AE 

[Since  AD  and  DX^^  are  commensurable.] 

4.  Take  now  as  unit  of  measure  \  AD.  This  measure  will 
be  contained  an  integral  number  of  times  in  AD  and  also  in 
i>Xi ;  further,  the  unit  of  measure  may  be  contained  once  in 
XiB,  leaving  a  remainder  X.^B  which  is  less  than  the  new 
unit  of  measure.  |  Draw  X2Y2  11  BC,  meeting  ^Oat  Y^. 

Then  DX^^^^Ii. 

AD      AE 

5.  Continue  in  this  manner  to  decrease  the  unit  of  measure 
indefinitely.  The  remainder  XB,  being  always  less  than  the 
unit  of  measure,  will  also  approach  0  as  limit. 


VARIABLES  AND   LIMITS 


Then    DX  =  BB,  amd  hence  ^  =  ^ 

AD     AD 


Also  EY=  EC,  and  hence 
6. 

7. 


EY^EC 
AE     AE 


263 

§  403,  (a) 
§  403,  (a) 


and are  variables  which  are  always  equal. 

AD  AE  .         ^      ^ 


DB^EG 
AD     AE 


403,  (5) 


Proposition  XVIII.     Theorem 

425.    Two  rectangles  having  equal  altitudes  are  to  each  other  as 
their  bases,  when  the  bases  are  incommensurable. 


Hypothesis.     Rectangles  ABCD  and  EFGH  have  equal  al- 
titudes AB  and  EF,  and  incommensurable  bases  BC  and  FG. 


Conclusion. 


EFGH^FG 
ABCD     BC' 


Proof.  1.  Divide  BC  into  any  number  of  equal  parts  (say 
two),  and  apply  one  of  these  parts  to  FG  as  unit  of  measure. 

2.  Since  BC  and  FG  are  incommensurable,  a  certain  num- 
ber of  segments  equal  to  the  unit  of  measure  will  extend  from 
Fto  Xi,  leaving  a  remainder  XiG  which  is  less  than  the  unit 
of  measure. 

3.  Draw  X^Yi±FG,  meeting  EH  at  Y^.  Then  rectangles 
EFXi  Yi  and  ABCD  have  equal  altitudes  and  commensurable 
bases. 

,  .   EFX,  Y,  ^  FXi 

'  "   ABCD       BC' 
Complete  the  proof. 
Suggestion.  —  Model  the  proof  after  that  for  §  424. 


264 


PLANE    GEOMETRY  —  BOOK  V 


Group  C.     Symmetry  in  Plane  Figures 

426.  Two  points  are  symmetrical  with  respect  to  a  third 
point,  called  the  Center  of  Symmetry,  when  the  latter  bisects 
the  segment  which  joins  them. 

Thus,  if  0  is  the  mid-point  of  segment  AB,  points  A  and  B  are  sym- 
metrical with  respect  to  0  as  center.  A O B 


427.  Two  points  are  symmetrical  with 
respect  to  a  straight  line,  called  the  Axis  of 
Symmetry,  when  the  latter  bisects  at  right 
angles  the  segment  which  joins  them.  c- 


B 


Thus,  if  CD  bisects  segment  AB  at  right  angles, 
points  A  and  B  are  symmetrical  with  respect  to 
CD  as  an  axis. 

428.  A  figure  is  symmetrical  with 
respect  to  a  center  when  every  straight 
line  drawn  through  the  center  cuts  the 
figure  in  two  points  which  are  symmetri- 
cal with  respect  to  that  center. 


429.  A  figure  is  symmetrical  with  respect  to 

an  axis  when  every  straight  line  perpendicular  /'^^V  j  7/^ 

to  the  axis  cuts  the  figure  in  two  points  which  ^-Mui//^^ 
are  symmetrical  with  respect  to  that  axis.  \fF^^ 

Ex.  102.     Does  a  circle  have  a  center  of  symmetry  ? 

Does  it  have  an  axis  of  symmetry  ? 

Does  it  have  more  than  one  axis  of  symmetry  ? 

Ex.  103.  (a)  Locate  upon  a  sheet  of  paper  a  point  0  and  four  other 
points  X,  r,  Z,  and  W. 

(b)  Construct  the  points  X',  Y',  Z',  and  W,  which  are  symmetrical 
respectively  to  X,  Y,  Z,  and  W,  with  respect  to  0  as  center. 

Ex.  104.  (a)  Draw  any  straight  line  AB  of  indefinite  length  and 
upon  one  side  of  it  locate  at  random  points  X,  Y,  and  Z. 

(b)  Construct  the  points  X,  Y',  and  Z',  which  are  respectively  sym- 
metrical to  X,  Y,  and  Z,  with  respect  to  AB  as  axis. 


SYMMETRY  265 

430.   Theorem.     Tioo  segments  wliicli  are  symmetrical  with 
respect  to  a  center  are  equal  and  parallel. 


.."  "O" 


Hypothesis.  Segments  AB  and  A'B'  are  symmetrical  with 
respect  to  center  0. 

Conclusion.     AB  and  A'B'  are  equal  and  parallel. 

Proof.  1.  Draw  lines  AA'  and  BB'  intersecting  at  0;  draw 
AB'  and  A'B.  §  428 

2.  0  bisects  AA'  and  BB'.  Prove  it. 

3.  .-.  AB'A'B  is  a  O.  Prove  it. 

4.  .'.  ^5  and  A'B'  are  equal  and  parallel. 

Ex.  105.     (a)  Draw  a  figure  something  like  the  adjoining 
one. 

(Let  ^B  and  FG  be  perpendicular  to  AG,  and  let  BCDEF 
be  a  curved  line. ) 

(6)  Construct  the  figure  symmetrical  to  ABCDEFG  with     e\ 
respect  to  u4(t  as  axis. 

F 

Ex.  106.  Prove  that  two  segments  which  are  equal  and  parallel  are 
symmetrical  with  respect  to  a  center. 

Ex.  107.  Prove  that  the  bisector  of  the  vertical  angle  of  an  isosceles 
triangle  is  an  axis  of  symmetry  of  the  triangle. 

Ex.  108.  How  many  axes  of  symmetry  does  an  equilateral  triangle 
have  ? 

Ex.  109.  Prove  that  the  intersection  of  the  diagonals  of  a  parallelo- 
gram is  the  center  of  symmetry  of  the  parallelogram. 

Ex.  110.     Does  a  rhombus  have  a  center  of  symmetry  ? 
Ex.  111.     Does  the  rhombus  have  an  axis  of  symmetry  ? 
Ex.  112.     Does  a  rectangle  have  an  axis  of  symmetry  ? 
Does  it  have  a  second  axis  of  symmetry  ? 
Does  it  have  a  center  of  symmetry  ? 


266 


PLANE   GEOMETRY  — BOOK  V 


431.  Theorem.  If  a  figure  is  symmetrical  with  respect  to 
each  of  two  perpendicular  axes,  it  is  symmetrical  with  respect  to 
their  intersection  as  center. 


c 

A 

/ 

/--, 

\p 

It 
D 

H 

> 

Q   / 

< 

0 

B 

o 

F 

Hypothesis.  Figure  AE  is  symmetrical  with  respect  to 
axes  XX'  and  YY\     XX'  is  perpendicular  to  YY'  at  0. 

Conclusion.     AE  is  symmetrical  with  respect  to  0  as  center. 

Proof.  1.  From  P'  any  point  of  AE  draw  P'Q  A.  to  XX'  at 
Q,  and  meeting  AE  again  at  P.     Then  PQ=  P'Q. 

2.  From  P  draw  PR±  YY'  meeting  YY'  at  R  and  meet- 
ing AE  again  at  P".     Then  PR  =  RP". 

3.  Draw  P'P". 

4.  YY'  II  PP'  and  bisects  PP". 

.-.  YY'  passes  through  the  mid-point  of  P'P".  Why  ? 

5.  Similarly,  XX'  passes  through  the  mid-point  of  P'P". 

6.  Hence  O,  the  intersection  of  XX'  and  YY',  must  be 
the  mid-point  of  P'P". 

7.  In  the  same  manner,  for  any  other  point  like  P'  of  AE 
there  is  a  corresponding  point  P"  of  AE,  such  that  P'P"  passes 
through  0  and  is  bisected  by  it. 

8.  .-.  AE  is  symmetrical  with  respect  to  0  as  center. 


Ex.  113. 

Ex.  114. 

in  Ex.  112. 

Ex.  115. 

Ex.  112. 

Ex.  116. 


Answer  for  a  square  the  questions  proposed  in  Ex.  112. 
Answer  for  an  isosceles  trapezoid  the  questions  proposed 

Answer  for  a  regular  hexagon  the  questions  proposed  in 

Answer  for  a  regular  pentagon  the  same  questions. 


MAXIMA  AND   MINIMA 


267 


Group  D.     Maxima  and  Minima  of  Figures 

432.  Two  figures  are  Isoperimetric  when  they  have  equal 
perimeters. 

433.  If  one  geometric  magnitude  of  a  number  which  satisfy 
certain  given  conditions  has  a  value  greater  than  that  of  any 
of  the  others,  it  is  called  the  Maximum ;  if  it  has  a  value  less 
than  that  of  any  of  the  others,  it  is  called  the  Minimum. 

Thus,  of  all  segments  drawn  from  a  given  point  to  a  given  line  the 
perpendicular  is  the  minimum  ;  again,  of  all  chords  of  a  circle^  the  diame- 
ter is  the  maximum. 

434.  Theorem.  Of  all  triangles  ivith  two  given  sides,  that  in 
which  these  sides  are  perpendicular  is  the  maximum. 


Hypothesis.     In  A  ABC  and  A  A'BC; 

AB  =  A'B ;  and  AB  ±  BC. 

Conclusion.     Area  of  A  ABC  >  area  of  A  A'BC. 
Proof.     1.  Draw  A'D  ±  BG, 

2.  A'B  >  A'D. 

3.  .-.  AB  >  A'D. 

4.  .-.  AB'BC>  A'D  '  BC 
[Multiplying  both  members  by  BG.'\ 

6.  But  area  A  ABC=  \  AB  •  BC, 

and  area  A  A'BC  =\A'D.BG. 
6.  .-.  area  A  ABC  >  area  A'BC 


Why? 
Why? 


Ex.  117.     Of  all  parallelograms  having  two  given  adjacent  sides,  that 
is  the  maximum  in  which  these  sides  include  a  right  angle. 


268 


PLANE   GEOMETRY  — BOOK  V 


435.   Theorem.     Of  isoperimetric  triangles  having  the  same 
base  J  that  which  is  isosceles  is  the  maximum. 


Ay 


--j:: 


r 


Hypothesis.     A  ABC  and   A  A!BQ  are   isoperimetric   and 
have  the  common  base  BQ\  A  ABQ  is  isosceles. 

Conclusion.     Area  of  A  ABQ  >  area  of  AA'BC. 

Analysis.     We  must  prove  the  altitude  AD  >  altitude  A'D/ 

Proof.     1.   Extend  BA  to  E,  making  AE  =  BA.     Draw  EO. 

2.  Z  BGE  is  a  rt.  Z,  for  it  can  be  inscribed  in  a  semicircle 
whose  center  is  A  and  whose  radius  is  AB. 

3.  Extend  J5;(7  to  X.     Btslw  A'E' =  A'C]  dvsiW  BE\     Con- 
struct A'F'  and  AF  both  perpendicular  to  EE'. 


4.  BA'  +  A'E'  =  BA'  -f  A'C. 

5.  BA  +  AE  =  BA-{-  AC. 

6.  '        But  BA'+ A'C  =  BA  + AC. 

7.  .-.  BA'  +  A'E'  ==BA  +  AE. 

8.  .-.  BA'  +  A'E'  =  SE-. 

9.  But  BA'  +  ^'^'  >  jB^'. 

10.  .-.  BE  >  J5^'. 

11.  «•.  (7^  >  (7J5;'. 

[Since  BC±EE',  and  5^  >  BE'.^ 

12.  .'.CF>CF'. 

[Since  Oi^=  i  0^,  and  C2^'  =  i  CE'.^ 

13.  .%  ^D  >  A'D'. 

(Prove  it.) 

14.  .-.  area  of  A  ABC  >  area  of  A  A'BC. 


Why? 

Why? 

Hyp. 

Why? 
Why? 
Why? 
Why? 


Why? 


MAXIMA  AND   MINIMA  .   269 

436.   Theorem.     Of  isoperimetric  polygons  having  the  same 
number  of  sideSy  the  maximum  is  equilateral. 


Hypothesis.  ABCDE  is  the  maximum  of  polygons  having 
the  given  perimeter  and  the  same  number  of  sides  as  ABCDE. 

Conclusion.  ABCDE  is  equilateral. 

Proof.  1.  Assume  that  AB  and  BC  are  unequal.  Draw 
AC. 

2.  Let  A  AB'C  be  the  isosceles  triangle  on  base  AC  having 
its  perimeter  equal  to  that  of  A  ABC. 

3.  Then  the  area  of  A  AB'C  >  area  of  A  ABC.  §  435 

4.  Then  the  area  of  AB'CDE  >  area  of  ABCDE. 

5.  But  this  is  impossible  for  ABCDE  is  the  maximum  of  all 
polygons  having  the  same  perimeter  and  the  same  number  of 
sides  as  ABCDE. 

6.  .-.  AB  and  BC  cannot  be  unequal. 

7.  Similarly  BC  =  CD  ^  DE,  etc. 

8.  .-.  ABCDE  is  equilateral. 

437.  Cor.  Of  all  isoperimetric  triangles,  the  maximum  is 
equilateral. 

Ex.  118.  A  parallelogram  and  a  rhombus  each  have  a  perimeter  of 
40  in.     Which  has  the  greater  area  ? 

Ex.  119.  A  man  is  planning  for  himself  a  house.  He  has  a  rectan- 
gular plan,  the  dimensions  of  which  are  30  ft.  and  20  ft.,  making  the 
perimeter  of  the  base  100  ft. 

Will  such  a  house  cover  a  greater  or  a  less  number  of  square  feet  than 
a  square  house  whose  perimeter  also  is  100  ft.  ? 


270  . 


PLANE   GEOMETRY  — BOOK  V 


438.   Theorem.     Of  isoperimetric  equilateral  polygons  of  the 
same  number  of  sides,  the  maximum  is  equiangular. 

H 


.'^D 


Fig.  1 


Fig.  2 


Fig.  3 


Hypothesis.  Consider  an  equilateral  polygon  of  which  AB, 
BC,  and  CD  are  any  three  consecutive  sides,  and  whose  re- 
maining part  is  denoted  by  P.  Assume  that  this  polygon  is 
the  maximum,  of  all  equilateral  polygons  isoperimetric  with 
the  given  polygon,  equal  in  area  to  it,  and  having  the  same 
number  of  sides  as  it. 

Conclusion.     This  polygon  is  equiangular. 
Plan.     We  shall  assume  that  Z  ABC  >  Z  BCD.     We  shall 
consider  the  three  cases :    Case  I.   AB  II  CD.      Case  II.   AB 
meets  CD  at  H.     Case  III.   AB  meets  CD  at  K. 

Proof.  Case  I.  (Fig.  1.)  1.  Let  E  be  the  mid-point  of  BC. 
Draw  EF  meeting  AB  prolonged  at  F,  making  EF  =  BE. 
Then  EF  extended  will  meet  CD  at  G. 

2.    Then  A  BEF  ^  A  ECG.  Prove  it. 

.-.  BF  =  CG,  and  EF  =  EG.  Why  ? 

.-.  EG  =  BC  Why  ? 

.-.  AB-\-BF+FG-\-GD=AB-\-BC+CG-\-GD.  Why? 
Hence  the  polygon  composed  of  AFGD  and  P  has  the 
same  perimeter  as  the  given  polygon,  composed  of  ABCD 
and  P. 

7.   Also  AFGD  and  ABCD  have  the  same  area.  Step  2 


MAXIMA  AND  MINIMA  271 

8.  Hence  the  polygon  composed  of  AFOD  and  P  has  the 
same  area  as  the  given  polygon,  composed  of  ABCD  and  P. 

9.  But  the  given  polygon  was  the  maximum  of  polygons 
having  the  given  number  of  sides.  Hence  the  polygon  com- 
posed of  AFGD  and  P  is  equal  to  the  maximum  of  polygons 
having  that  number  of  sides. 

10.  Hence  the  polygon  composed  of  AFOD  and  P  is 
equilateral.  §  436 

11.  But  this  is  impossible  since  AF  >  DO. 
•    12.   Hence  Z  ABC  cannot  be  >  Z  BCD. 

Case  II.     (Fig.  2.)     Assume  that  AB  meets  CD  at  H, 

1.  Let  HE  bisect  Z  BHC,  meeting  BC  at  E. 

2.  Revolve  A  BCH  on  HE  as  axis  until  it  takes  the  position 
oiAFGH. 

3.  Then  FG=BC\  BF=  CO ;  and  A  BEF=A  CEO.    Why  ? 

4.  .'.  AB+BF-\-FG-\-GD=AB-{-BC-^CG-hGD.        Why? 
Complete  the  proof  as  in  steps  6  to  12  inclusive,  of  Case  I. 

Case  III.     (Fig.  3.)     Assume  that  AB  and  CD  meet  at  K. 

1.  Let  KE  bisect  Z  BKC,  meeting  BC  at  E. 

2.  Revolve  A  BCK  on  KE  as  axis  until  it  takes  the  position 
of  A  FOK 

3.  ThenFO^BC;  BF=CO',  2indABEF=ACEO.    Why? 

4.  .:  AB-^BF-^FG-\-GD=AB  +  BC-i-CO+GD.        Why? 
Complete  the  proof  as  in  steps  6  to  12  inclusive,  of  Case  I. 

It  follows  from  Cases  I,  II,  and  III  that  Z  ABC  cannot  be 
>  Z  BCD. 

In  the  same  manner,  it  can  be  proved  Z  ABC  cannot  be 
<  Z  BCD. 

Hence  Z  ABC  =  Z  BCD. 

Since  these  are  any  two  consecutive  angles  of  the  given 
polygon,  then  the  given  polygon  must  be  equiangular. 

439.  Cor.  Of  isoperimetric  polygons  having  the  same  num- 
ber of  sides,  the  maximum  is  regular.  (§  436  and  §  437) 


272 


PLANE  GEOMETRY  — BOOK  V 


440.   Theorem.     Of  two  isoperimetric  regular  polygons,  that 
which  has  the  greater  number  of  sides  has  the  greater  area. 


Hypothesis.     ABQ  is  an  equilateral  triangle,  and  M  is  an 
isoperimetric  square. 

Conclusion.     Area  of  Jf  >  area  of  A  ABC. 

Proof.     1.   Let  D  be  any  point  in  side  AB  of  A  ABC. 

2.  Draw  DO,  and  construct  upon  it  as  base  isosceles  A  CDE 
isoperimetric  with  A  BCD. 

3.  Area  of  A  CDE  >  area  of  A  BCD. 

4.  .-.  area  of  ADEC  >  area  of  A  ABC 

5.  But  ADEC  and  square  M  are  isoperimetric,  and  hence 
area  of  Jf  >  area  of  ADEC  §  439 

6.  .-.  area  of  M>  area  of  A  ABC 

In  like  manner  it  can  be  proved  that  the  area  of  a  regular 
pentagon  is  greater  than  that  of  an  isoperimetric  square ;  etc. 

441.   Cor.     The  area  of  a  circle  is  greater  than  the  area  of 
any  polygon  having  an  equal  perimeter. 


SUPPLEMENTARY  EXERCISES  273 

SUPPLEMENTARY  EXERCISES 
BOOK  I 

Ex.  1.  Two  quadrilaterals  are  congruent  if  three  sides  and  the  two 
included  angles  of  one  are  equal  respectively  to  three  sides  and  the  two 
included  angles  of  the  other. 

Suggestion.  — Prove  by  superposition. 

Ex.  2.  Two  quadrilaterals  are  congruent  if  three  angles  and  the  two 
included  sides  of  one  are  equal  respectively  to  three  angles  and  the  two 
included  sides  of  the  other. 

Ex,  3.  Prove  that  the  base  angles  of  an  isosceles  triangle  are  equal, 
using  the  following  construction. 

Hypothesis.  AB  =  AC. 

Conclusion.  ZABC  =  ZA  CB. 

Construction.     Extend  AB  to  D.     Extend  AC  to 
E,  making  CE  =  BD.     Draw  DC  and  BE. 

Plan.     1.    Prove  A  ADC  ^  A  ABE  in  order  to 
prove  DC=  BE. 

2.  Prove  ADBC^ABCE. 

3.  Prove  ZDBC  =  ZBCE. 

4.  Prove  ZABC=ZACB. 

Ex.  4.  If  AB  and  ^C  are  two  equal  chords  of  the  circle  whose  center 
is  O,  then  the  radius  OA  bisects  ZBAC. 

Ex.  5.  Books  for  carpenters  give  the  follow- 
ing method  of  bisecting  an  angle  by  means  of 
the  "  square  "  alone. 

Make  OD  and  OC  of  equal  length.  Place 
the  square  so  that  DP=  CP.  Then  OP  bisects 
ZAOB. 

Prove  that  the  method  is  correct. 

Ex.  6.  Prove  that  the  bisectors  of  homologous  angles  of  congruent 
triangles  are  equal. 

Suggestions.  —  1.    Recall  §  66. 

2.  Remember  that  the  homologous  sides  and  angles  of  two  congruent 
triangles  are  equal. 

Ex.  7.     Construct  the  angle  which  is  double  ZJ5  of  Ex.  61,  Book  I. 

Ex.  8.     Construct  the  angle  which  is  the  sum  of  Z  ^  and  Z  B  of  Ex.  61. 

Ex.  9.  Construct  an  isosceles  triangle  having  its  equal  sides  3  in.  in 
length  and  the  angle  included  by  them  equal  to  ZB  given  in  Ex.  61, 
Book  I. 


274 


SUPPLEMENTARY  EXERCISES 


Ex.  10.  If  a  diagonal  of  a  quadrilateral  ABCD  bi- 
sects two  of  its  angles,  it  is  perpendicular  to  the  other 
diagonal  and  bisects  it. 

Suggestions.  —  1.  Let  AC  bisect  Z.A  and  ^C;  prove 
AC  1  BD  and  AC  bisects  BD. 

2.   Try  to  prove  AB  =  AD  and  BC=DC. 

Ex.  11.   In  the   adjoining  figure,  if   J.0,    BO, 

and  CO  are  extended  to  Z,  F,  and  X  respectively, 

so  that  AO=OZ,  BO=OY,  and  C0=  OX,  then 

CiABC^/\ZYX. 

(First  prove  AB  =  ZY,  BC  =  XY,  and  AC=XZ.) 

After  proving  /\ABC  '^  A  XYZ,  vi^hat  angle  does  /.BCA  equal  ? 

Ex.  12.  Prove  that  homologous  medians  of  congruent  triangles  are 
equal. 

Suggestion.  —  Read  the  suggestions  for  Ex.  6,  p.  273. 

Ex.  13.  If  two  triangles  have  two  sides  and  the  median  to  one  of 
them  equal  respectively  to  two  sides  and  the  corresponding  median  of  the 
other,  the  triangles  are  congruent. 

Suggestion. — Read  the  note  following  §  77. 

Ex.  14.  Construct  the  perpendicular-bisector  of  a  segment  taken 
along  the  lower  edge  of  the  paper. 

Ex.  15.  Draw  any  angle  and  construct  its  bisector.  Through  its 
vertex  construct  a  line  perpendicular  to  the  bisector.  Prove  that  this  last 
line  makes  equal  angles  with  the  sides  of  the  given  angle. 

Ex.  16.  Construct  a  line  through  a  given  point 
within  a  given  acute  angle,  which  will  form  with  the 
sides  of  the  angle  an  isosceles  triangle. 

Suggestion.  —  If  A  ABC  represents  the  desired  triangle, 
and  AH  bisects  Z  CAB,  then  CB  1  AH.  Try  now  to  work 
toward  this  figure  if  only  I. FAG  and  point  D  within  it 
are  given. 

Ex.  17.     Prove  Cor.  1  (§  96)  if  Z  3  =  Z  7. 

Ex.  18.  Prove  Cor.  3  (§98)  ifZ3  +  Z5  = 
1  St.  Z. 

Ex.  19.  Prove  that  AB  \\  CD  (Fig.  §  98)  if 
Z4  +  Z7  =  lst.  Z. 

Ex.  20.  If  ^5  =  CZ)  and  Z  1  =  Z  2  in  the  ad- 
joining figure,  prove  ABW  CD  and  also  AGW  BD. 

Suggestion.  —  UecSiW  §  95. 


SUPPLEMENTARY   EXERCISES 


275 


tread -S 

A<^^ 

y^ 

tread 

< 

^ 

1 

Ex.  21.  In  building  stairs  two  or  more 
stringers  are  required.  To  make  a  stringer 
having  a  9-in.  tread  and  a  6-in.  riser,  a  car- 
penter uses  his  square  as  in  the  figure  below. 
For  each  step  he  places  his  square  so  that  the 
9-in.  mark  and  the  0-in.  mark  fall  along  the 
edge  of  the  board  from  which  he  is  cutting 
the  stringer. 

Will  the  treads  all  be  parallel  ?    Why  ? 

Will  the  risers  all  be  parallel  ?    Why  ? 

Will  each  riser  be  perpendicular  to  its  tread  ? 
Why? 

Ex.  22.  One  triangle  used  by  draughtsmen  has  an  angle  of  90°  and 
an  angle  of  GO"".     Why  should  it  be  called  a  "  60-30  "  triangle  ? 

Ex.  23.  The  other  triangle  used  by  draughtsmen  has  an  angle  of  90=^ 
and  an  angle  of  45°.     How  large  is  the  remaining  angle  ? 

Ex.  24.  In  a  A  ABC,  it  ZA  =  90°,  and  Z B  =  Z  C,  how  large  are 
ZB  and  ZC? 

Ex.  25.  Find  the  three  angles  of  a  triangle  if  the  second  is  four  times 
the  first,  and  the  third  is  seven  times  the  first.     (Algebraic  solution.) 

Ex.  26.  Find  the  three  angles  of  a  triangle  if  the  second  exceeds  the 
first  by  40°,  and  the  third  exceeds  the  second  by  40°. 

Ex.  27.  The  vertical  angle  of  an  isosceles  triangle  is  n  degrees.  Ex- 
press each  of  the  base  angles. 

Ex.  28.  One  base  angle  of  an  isosceles  triangle  is  n  degrees.  Ex- 
press each  of  the  other  angles  of  the  triangle. 

Ex.  29.  Determine  by  construction  the  angle  C  of  a  A  ABC  if  ZA 
and  Z  B  are  the  angles  given  in  Ex.  61,  Book  I. 

Ex.  30.  Prove  that  two  isosceles  triangles  are  congruent  when  the 
vertical  angle  and  the  base  of  one  are  equal  respectively  to  the  vertical 
angle  and  the  base  of  the  other. 

Suggestion.  —  Prove  the  homologous  base  angles  also  are  equal. 

Ex.  31.  If  one  acute  angle  of  a  right  triangle  is  35°,  how  large  is  the 
other  acute  angle  ? 

Ex.  32.  If  perpendiculars  be  drawn  from  any  point  in  the  base  of  an 
isosceles  triangle  to  the  equal  sides,  they  make  equal  angles  with  the  base. 

Suggestion.  —The  proof  i.s  based  on  §  37  and  §  109. 

Ex.  33.  Prove  that  the  altitude  drawn  to  the  hypotenuse  of  a  right 
triangle  divides  the  right  angle  into  two  parts  which  are  equal  respectively 
to  the  acute  angles  of  the  right  triangle. 


276  SUPPLEMENTARY  EXERCISES 

Ex.  34.  If  two  opposite  angles  of  a  quadrilateral  are  equal  and  if  the 
diagonal  joining  the  other  two  angles  bisects  one  of  them,  then  it  bisects 
the  other  also. 

Ex.  35.  If  two  triangles  have  two  angles  and  the  bisector  of  one  of 
these  angles  equal  respectively  to  two  angles  and  the  corresponding  bisector 
of  the  other,  the  triangles  are  congruent. 

Suggestion.  — Recall  the  note  following  §  77. 

Ex.  36.  If  two  triangles  have  two  sides  and  the  altitude  drawn  to 
one  of  them  equal  respectively  to  two  sides  and  the  corresponding  altitude 
of  the  other,  the  triangles  are  congruent. 

Suggestion.  —  Read  the  note  following  §  77. 

Ex.  37.     Construct  a  pattern  for  the  pointed  end  of 

a  belt,  assuming  that  the  belt  material  is  2  in.  wide,  and  ^  '2'       kj^ 

that  the  point  is  to  project  1  in.  beyond  the  square  end  '^ 

of  the  belt. 

Ex.  38.  Draw  any  straight  line  of  indefinite  length  and  select  two 
points  not  in  it.  Find  the  point  in  the  line  which  is  equidistant  from  the 
two  given  points. 

Ex.  39.  Find  a  point  in  one  side  of  a  triangle  which  is  equidistant 
from  the  other  two  sides  of  the  triangle.  ^^ 

Ex.  40.  Prove  that  either  exterior  angle  at  the 
base  of  an  isosceles  triangle  is  equal  to  the  sum  of  a 
right  angle  and  one  half  the  vertical  angle.  A^ 

E- 

Ex.  41.  If  from  the  vertex  of  one  of  the  equal  angles 
of  an  isosceles  triangle  a  perpendicular  be  drawn  to  the 
opposite  side,  it  makes  with  the  base  an  angle  equal  to 
one  half  the  vertical  angle  of  the  triangle. 

Suggestion.  —  Construct  the  bisector  of  Z  C. 

Ex.  42.  A  ABC  is  an  equilateral  triangle.  BP,  the  bisector  of  Z  B, 
meets  ^C at  P;  CM,  the  bisector  of  exterior  angle  ACB,  meets  BP 
extended  at  M.     MNis  perpendicular  to  CB.    Prove  MN  =  BP. 

Ex.  43.  If  the  equal  sides  of  an  isosceles  triangle  be  extended  be- 
yond the  base,  the  bisectors  of  the  exterior  angles  so  formed  form  with 
the  base  another  isosceles  triangle. 

Ex.  44.  If  A  ABC  and  A  ABD  are  two  tri- 
angles on  the  same  base  and  on  the  same  side  of 
it,  such  that  AC  =  BD  and  AD  =  BC,  and  AD 
and  BC  intersect  at  0,  then  A  OAB  is  isosceles. 


SUPPLEMENTARY   EXERCISES 


277 


Ex.  45.  If  CD  is  the  bisector  of  Z  C  of  A  ABC, 
and  DF  be  drawn  parallel  to  ^O  meeting  BC  2it  E 
and  the  bisector  of  the  angle  exterior  to  C  at  F, 
prove  DE-EF.  B- 

Sugrjestion.  —  Compare  BE  and  EF  with  EC.  .^ 

Ex.  46.  If  equiangular  triangles  be  constructed  upon  the  sides  of  any 
triangle,  the  lines  drawn  from  their  outer  vertices  to  the  opposite  vertices 
of  the  given  triangle  are  equal. 

Suggestion.  — Recall  §  124.  ^D 

Ex.  47.  If  -4C  be  drawn  from  the  vertex  of  the  right 
angle  to  the  hypotenuse  of  right  A  BCD  so  as  to  make 
Z  ACD  =  ZZ>,  it  bisects  the  hypotenuse. 

Suggestion.  — Prove  ^B  =  Z.ACB  hy  ^  109  and  §  37. 

Ex.  48.  If  the  angle  at  the  vertex  of  isosceles 
A  ABC  is  equal  to  twice  the  sum  of  the  equal  angles 
B  and  C,  and  if  CD  is  perpendicular  to  BC,  meeting 
BA  extended  at  2),  prove  A  ACD  is  equilateral. 

Suggestion.  —  Determine  the  number  of  degrees  in  each  angle  of  A  ABC. 

aC 

Ex.  49.  If  the  bisectors  of  the  equal  angles  of  an 
isosceles  triangle  meet  the  equal  sides  at  D  and  E  re- 
spectively, prove  that  DE  is  parallel  to  the  base  of  the 
triangle.  a^ :ab 

Suggestions.  —  1.    Compare  Z  CED  +  L  CDE  with  Z  ^  +  Z  ^  (§  106) . 

2.  Is  LCED=^CDE?    3.  Is  ZC£Z>  =  Z^? 

Ex.  50.  Prove  that  two  parallelograms  are  congruent  if  two  sides 
and  the  included  angle  of  one  are  equal  respectively  to  two  sides  and  the 
included  angle  of  the  other. 

Suggestion.  —  Prove  by  superposition.    Recall  §  132. 

Ex.  51.  Prove  that  the  sum  of  the  perpendiculars 
drawn  from  any  point  within  an  equilateral  triangle  to 
the  sides  of  the  triangle  is  equal  to  the  altitude  of  the 
triangle. 

Prove  OR  -\-  0F+  0D  =  BG. 

Suggestions.  —  1.   Let  KM  be  II  ^C  and  KE  1  AB. 

2.  Compare  EK  and  BL.    3.  Prove  0/J  +  0F=  EX. 

Ex.  52.   An  ironing  board  is  supported  on  each  "^ 
side  as  shown  in  the  adjoining  figure.    If  ^0  =  OB 
and  DO  =  OC,  prove  that  AC  is  always  parallel  to 
the  floor  DB.  " 


G  R 


278 


SUPPLEMENTARY  EXERCISES 


Ex.  53.  What  angle  is  formed  by  the  bisectors  of  two  consecutive 
angles  of :  (a)  a  rectangle  ?  (6)  an  equilateral  triangle  ?  (c)  a  parallel- 
ogram ?  B^ _,C 

Ex.  54.  Prove  that  the  bisectors  of  the  in- 
terior angles  of  a  parallelogram  form  a  rectangle. 


Ex.  55.  Construct  a  rhombus  whose  sides  are  each  3  in.  and  whose 
acute  angles  are  each  45°.     Draw  and  measure  its  diagonals. 

Ex.  56.    Construct  a  rhombus,  having  given  one  side  and  one  diagonal. 

Ex.  57.   Prove  that  the  two  altitudes  of  a  rhombus  are  equal. 

Ex.  58.  If  on  the  diagonal  BB  of  square  ABCD  a  distance  BE  is 
taken  equal  to  AB,  and  if  EF  is  drawn  perpendicular  to  BD  meeting  AD 
at  F,  then  AF  =  EF  =  ED. 

Suggestion.  —  What  kind  of  angle  is  angle  EDF? 

Ex.  59.  If  AD  and  BD  are  the  bisectors  of  the  exterior  angles  at  the 
ends  of  the  hypotenuse  AB  of  right  triangle  ABC,  and  DE  and  Z>jPare 
perpendicular  respectively  to  CA  and  CB  extended,  prove  CEDE  is  a 
square. 

Suggestion. —Ilecall  §  143.    Prove  DE  =  DF,  using  §  120,  I. 

Ex.  60.     Prove  that  the  bisectors  of  the  angles  of  ^ 

a  rectangle  form  a  square. 

Suggestions.  —  1.   Make  a  plan  based  upon  §  143. 

2.   To  prove  EF  =  EH,  prove  AF=BH  and  AE  =     ^^ 


BE. 

Ex.  61.    If  the  non -parallel  sides  of  an  isosceles  trapezoid  are  extended 
until  they  meet,  they  form  with  the  base  an  isosceles  triangle. 

Ex.  62.   If  the  line  joining  the  mid-points  of  the  bases  of  a  trapezoid  is 
perpendicular  to  the  bases,  the  trapezoid  is  isosceles. 

Ex.  63.  If  the  bisectors  of  the  interior  angles  of  a 
trapezoid  do  not  meet  at  a  point,  they  form  a  quad- 
rilateral, two  of  whose  angles  are  right  angles. 

Suggestion.— Frove  Z.FEH  and  ^FGH  are  right  A' 
angles. 

Ex.  64.  If  D  is  the  mid-point  of  side  ^C  of  isosceles 
A  ABC,  and  DE  is  perpendicular  to  base  BC,  then  EC 
is  \  BC. 

Suggestion.  — Draw  DF  parallel  to  AB. 


SUPPLEMENTARY   EXERCISES  279 

Ex.  65.  A  BCD  is  a  trapezoid  whose  parallel 
sides  AD  and  BC  are  perpendicular  to  CD.  If  Eis 
the  inid-point  of  AB,  prove  EC  =  ED. 

Suggestion.  — Br&w   EF   parallel    to    AD.    Recall    A 
§  149. 

Ex.  66.  The  following  method  of  dividing  a  segment  into  equal  seg- 
ments may  be  used. 

To  divide  AB  into  five  equal  parts.  "H,,—^''^ 

1.  Draw  ^ C  making  with  ^5  any  con-  X-^'X''' \        \ 
venient  angle.                                                     a.--''^"    \       \       \     ,}-^ 

2.  Draw  BD  parallel  to  AC.  \        \       \  ^,,^—'''0" 
8.   Lay  off  on  40  five  equal  segments,        \ ^ "■-){""  ^ 

and  on  BD  five  other  segments  of  the  same         ^ 
length. 

4.    Connect  the  points  of  division  as  in  the  figure. 

Prove  now  that  AB  is  divided  into  five  equal  segments. 

Ex.  67.  If  the  base  of  an  isosceles  triangle  be  trisected,  the  lines 
joining  the  points  of  trisection  to  the  vertex  of  the  triangle  are 
equal. 

Ex.  68.  Prove  that  the  line  which  joins  the  mid-points  of  two  sides  of 
a  triangle  bisects  any  segment  drawn  to  the  third  side  from  the  opposite 
vertex. 

Ex.  69.     If  E  and  F  are  the  mid-points  of  BC  and  AD  respectively 
of  parallelogram  ABCD,  prove  that  AE  and  CF 
trisect  BD. 


Suggestion.  — Frove  AE  ||  FC,  by  proving  AECF  /     ^^ 

is  a  parallelogram.    Then  prove  that  AE  bisects  BH    a 
aud  CF  bisects  GB. 

Ex.  70.  If  E  and  i?'are  the  mid-points  of  sides  AB  and  AC  respec- 
tively of  t^ABC,  and  AD  is  the  perpendicular  from  A\.o  BC^  prove 
/.EDF^AEAF. 

Suggestion.  —Recall  Ex.  175,  Book  I. 

Ex.  71.  If  ^and  G  are  the  mid-points  of  AB 
and  CD  respectively  of  quadrilateral  ABCD,  and 
/rand  L  are  the  mid-points  of  diagonals  AC  and 
BD  respectively,  prove  that  EKGL  is  a  parallelo- 
gram. 

Suggestion.  — ^cdkW  §  161. 


280 


SUPPLEMENTARY  EXERCISES 


Ex.  72.  Prove  that  the  lines  joining  the 
mid-points  of  the  opposite  sides  of  a  quadri- 
lateral and  the  line  joining  the  mid-points 
of  the  diagonals  of  the  quadrilateral  meet  in 
a  point. 

Suggestion.  —  Recall  Ex.  71. 

Ex.  73.     Prove  that  the  line  joining  the  ^ 

mid-points  of  the  diagonals  of  a  trapezoid  is  parallel  to  the  bases  and 
equal  to  |  their  difference. 

Suggestions. —  1.  Dravr  EG  II  AD  meeting  CD  at  G. 

2.  Prove  EG  passes  through  point  F. 

3.  Compare  EG  with  AD  and  FG  with  BC. 

Ex.  74.  If  the  perpendiculars  AE,  BF,  CG, 
and  DH  be  drawn  from  the  vertices  of  parallelo- 
gram ABCD  to  any  line  in  its  plane  not  intersecting 
its  surface,  prove  that  AE  +  CG  =  BF  -»-  DH. 

Suggestion.  —  See  adjoining  figure.    Apply  §  153. 

Ex.  75.     Prove  that  the  sum  of  any  three  sides  of  a  quadrilateral  is 
greater  thau  the  fourth  side. 
Suggestion.  —  Draw  a  diagonal. 

Ex,  76.     Prove  that  the  sum  of  the  lines  drawn  from  any  point  within 
a  triangle  to  the  vertices  is  less  than  the  sum  of  the  three  sides. 
Suggestion.  —  1.   Let  0  within  A  ABC  be  joined  to  A,  B,  and  C 

2.  OA-\-OB<AC+BC.     (Ex.  188,  Book  I.) 

3.  Similarly  express  OB -\-  OC'and  also  OC -\-  OA. 

4.  Add  these  inequalities  and  divide  by  2. 

Ex.  77.     In  triangle  ABC,  if  D  is  any  point  on  ^C  so  that  AD  =  AB, 
then  BC>  DC. 

Suggestion.  —  Compare  BC-\r  AB  with  AC. 

Ex.  78.     Prove  that  each  of  the  equal  sides  of  an  isosceles  triangle  is, 
greater  than  one  half  the  base. 

Ex.  79.     If  0  is  any  point  within  triangle  ABC,  then  AO-\-BO-\-GO 
>  \  perimeter. 

Suggestions.  — 1.   Apply  §  159  (a)  to  each  side  of  the  triangle. 
2.   Add  the  inequalities  and  divide  by  2. 

Ex.  80.     Prove  that  any  side  of  a  triangle  is  less  than  one  half  the 
perimeter  of  the  triangle. 

Suggestions.  —  1.  Apply  §  159  (a)  to  one  side. 

2.  Add  that  side  to  both  members  of  the  inequality. 


SUPPLEMENTARY   EXERCISES 


281 


Ex.  81.  Prove  that  the  median  to  any  side  of  a  triangle  is  l^s  than 
one  half  the  perimeter  of  the  triangle. 

Suggestion.  — The  median  lies  in  each  of  two  As. 

Apply  §  159  (a)  to  the  median  in  each  A,  and  add. 

Ex.  82.  Prove  that  the  median  to  any  side  of  a  triangle  is  less  than 
one  half  the  sum  of  the  other  two  sides  of  the  triangle. 

Suggestion.  —  1.  Extend  the  median  its  own  length,  through  the  side  of  the 
triangle.  Connect  the  end  of  the  new  segment  with  one  of  the  other  vertices 
of  the  triangle. 

Ex.  83.  Prove  that  the  median  to  any  side  of  a  triangle  is  greater 
than  one  half  the  sum  of  the  other  two  sides  diminished  by  the*  side  to 
which  it  is  drawn. 

Ex.  84.  Prove  that  the  sum  of  the  medians  to  the  sides  of  a  triangle 
is  greater  than  one  half  the  perimeter  of  the  triangle. 

Suggestion. —  Apply  Ex.  83  to  each  of  the  medians. 


Ex.  85.  The  bisectors  of  the  exterior  angles  at 
two  vertices,  and  the  bisector  of  the  interior  angle  at 
the  third  vertex  of  a  triangle  are  concurrent. 

Suggestion.  —  The  proof  is  like  that  for  §  169. 


Ex.  86.     If  two  medians  of  a  triangle  are  equal, 
isosceles. 

Ex.  87.  Prove  that  the  line  joining  the  ortho- 
center  of  a  triangle  to  the  circum-center  of  the  tri- 
angle passes  through  the  center  of  gravity  (§178)  of 
the  triangle. 

Suggestions.  —  1.  Draw  AR,  and  try  to  prove  that  K 
is  the  center  of  gravity,  by  proving  that  AK  =  2  KR. 
.      2.  Recall  §  152,  and  Ex.  190,  Book  I. 

Ex.  88.  If  0  is  the  point  of  intersection  of  the 
medians  AD  and  BE  of  equilateral  triangle  ABC,  and 
OF  is  drawn  parallel  to  AC,  meeting  BC  at  F,  prove 
that  DF  ial  BC. 

Suggestion.  —  Let  G  be  the  mid-point  of  OA,  and 
draw  GH  ||  AC;  also  imagine  a  line  through  D  II  AC. 
Apply  §  147. 

Ex.  89.     If  the  exterior  angles  at  the  vertices  A  and  J5  of  A  ABC  are 
bisected  by  lines  which  meet  at  D,  prove  Z  D  =  ^  /IB -\- ^ZA. 


282 


SUPPLEMENTARY  EXERCISES 


Proof.   1.    ZD  =  lSO- ZDAB- ZABD.        Why? 

2.  ZDAB=lZEAB  =  l(iZC+ZABG).  Why? 

3.  Similarly  Z  ABD  =  ? 

4.  180°  =  ZBAC  +  ZC  +  ZABG.      Why? 

5.  Substitute  in  step  1,  and  complete  the  proof. 
Note. — This  proof  is  typical  of  many  that  involve"  ^^ 

numerical  relations  among  angles  of  a  figure.     In  triangles,  the  facts  in 
§§  106,  109,  and  110  are  used  frequently. 

Ex.  90.    Prove  that  the  exterior  angle  at  the  base  of  an  isosceles 
triangle  equals  the  angle  between  the  bisectors  of  the  base  angles. 

Ex.  91.  D  is  any  point  in  the  base  BG  ol  isosceles 
triangle  ABG.  The  side  ^C  is  extended  from  G  to  E, 
so  that  GE  equals  CD,  and  DE  is  drawn,  meeting  AB 
at  F.     Prove  Z  AFE  =  3  Z  AEF. 

Suggestions.  —  l.Z  AFE  is  exterior  to  A  BDF. 

2.   ZB  =  ZA  CD,  which  is  exterior  to  A  ODE. 

Ex.  92,  If  GD  is  the  altitude  to  the  hypotenuse 
AB  of  right  triangle  ABC^  and  E  is  the  mid-point  of 
AB,  prove  ZDCE=ZA-ZB. 

Suggestions.  —  1.   ZDCE  is  the  complement  of  ZDEC.  ^ 
Why  ? 

2.  Express  Z  DEC.  3.  Recall  Ex.  175,  Book  I. 

Ex.  93.  If  GD  is  the  altitude  to  the  hypotenuse 
AB  of  right  triangle  ABC,  and  GE  is  the  bisector  of 
Z  C,  meeting  AB  at  E,  then  ZDGE=\  (ZA-  ZB). 

Suggestions.  —  1.   ZDCE  =  Z  ACE  -  ZACD. 

2.   ZACE=^h90°.  3.  90°=  ZA-\-ZB 

Ex.  94.     If  ZB  oi  A  ABG  is  greater  than  Z  C,  and  BD  is  drawn  to 
AG  making  AD  equal  to  AB,  prove 

Z  ADB  =  1{ZB  +  Z  G),  and  Z  GBD  =  l(ZB-Z<y). 

Suggestions.  — 1.  First  apply  §  110. 

2.   ZDBC=  ZB-ZABD=  ZB- ZADB.  Why? 

Ex.  95.    If  Z>  and  E  are  the  mid-points  of  sides  q 
BG  and  AG  respectively,  of  l^ABG,  and  AD  hQ 
extended  to  F  and  BE  to  O,  making  DF  -.  AD 
and  EG  =  BE,  prove  that  GGF  is  a  straight  line 
and  that  GG  =  GF. 

Suggestion.  — 'Recsill  Ex.  146,  Book  I,  and  §  90. 


Why? 
Why? 


C 

'^N 

\ 

A 

\ 

'w 

/  ^ 
>< 

><D      / 

A 

SUPPLEMENTARY  EXERCISES 


283 


Ex.  96.  If  the  median  drawn  from  any  vertex  of  a  triangle  is  greater 
than,  equal  to,  or  less  than  one  half  the  opposite  side,  the  angle  at  the 
vertex  is  acute,  right,  or  obtuse  respectively.     (§  161.) 

Ex.  97.  The  perpendicular  from  the  inter- 
section of  the  medians  of  a  triangle  to  any  straight 
line  in  the  plane  of  the  triangle,  not  intersecting 
its  surface,  is  equal  to  one  third  the  sum  of  the 
perpendiculars  from  the  vertices  of  the  triangle  to 
the  same  line.     (§  163.) 

G     Q  K    L      M  H 

Ex.  98.  Prove  that  the  diagonals  of  an  oblique-angled  parallelogram 
are  unequal,  the  one  joining  the  acute  angles  being  the  greater. 

Ex.    99.     Define: 

(a)  parallelogram  (/)  isosceles  trapezoid 

(b)  rectangle  (^g)  altitude  of  a  parallelogram 

(c)  square  (h)  altitude  of  a  trapezoid 

(d)  rhombus  (i)   median  of  a  trapezoid 

(e)  trapezoid 

Ex.  100.  What  are  the  important  facts  known  about  every  parallelo- 
gram ? 

Ex.  101.  State  four  theorems  by  which  a  quadrilateral  can  be  proved 
a  parallelogram. 

Ex.    102.  State  facts  known  about  a  rectangle. 

Ex.    103.  State  facts  known  about  a  square. 

Ex.    104.  State  facts  known  about  a  trapezoid. 

Ex.   105.  State  facts  known  about  an  isosceles  trapezoid. 

Ex.    106.  State  methods  for  proving  two  segments  equal. 

Ex.    107.  State  methods  for  proving  two  angles  equal. 

Ex.   108.  State  methods  for  proving  two  lines  are  parallel. 


BOOK  II 

Ex.  1.     If  AB  is  one  of  the  non-parallel  sides  of  a  trapezoid  circum- 
scribed about  a  circle  whose  center  is  0,  prove  ZAOB  is  a  right  angle. 
Suggestion.  —  Recall  Ex.  38  (6) ,  p.  106. 

Ex.  2.  The  straight  line  joining  the  raid-points  of  th^  non-parallel 
sides  of  a  circumscribed  trapezoid  is  equal  to  \  the  perimeter  of  the 
trapezoid. 

Suggestion.  —  Recall  Ex.  39,  p.  106. 


284 


SUPPLEMENTARY   EXERCISES 


Ex.  3.  If  tangents  are  drawn  to  a  circle  at  the  extremities  of  any  pair 
of  diameters  which  are  not  perpendicular  to  each  other,  the  figure  formed 
is  a  rhombus.     (Recall  Ex.  39,  p.  106.) 

Ex.  4.  If  the  angles  of  a  circumscribed  quadrilateral  are  right  angles, 
the  figure  is  a  square. 

Ex.  5.  A,  B,  C,  and  Z>  are  four  points  in  a 
straight  line,  B  lying  between  C  and  D ;  EF  is  a 
common  tangent  to  the  circles  drawn  upon  AB  and 
CD  as  diameters.     Prove  Z  BAE  =  Z  DCF. 

Ex.  6.  If  ABCD  is  a  quadrilateral  circumscribed 
about  a  circle  whose  center  is  0,  prove  that  ZAOB  + 
Z  COD  =  180°. 

SuggesHo7i.~Com]^8iTe  ZEOB  and  ZBOF;  LEO  A 
and  ZAOH,  etc. 


Ex.  7.  Construct  a  figure  like  the  one  adjoin- 
ing, using  for  the  equal  circles  from  which  it  is 
constructed  the  radius  1  in.  Notice  that  the  cir- 
cles are  tangent  circles. 


Ex.  8.  A  very  small  triangular 
piece  of  ground  AB  G  lies  in  the  inter- 
section of  three  streets.  Make  a 
drawing  to  scale  (1"  =20').  Then 
construct  corners  which  will  be  both 
more  useful  and  more  artistic  than 
the  sharp  corners.  Indicate  on 
your  drawing  the  radii  of  the  circles 
you  construct  in  the  corners. 

Ex.  9.  Prove  that  an  inscribed  angle  whose  intercepted  arc  is  less 
than  a  semicircle  is  an  acute  angle  ;  and  one  whose  arc  is  greater  than  a 
semicircle  is  an  obtuse  angle. 

Ex.  10.  If  any  number  of  equal  angles  are  inscribed  in  an  arc,  their 
bisectors  pass  through  a  common  point. 

Ex.  11.  If  the  diagonals  of  an  inscribed  quadrilateral  intersect  at  the 
center  of  the  circle,  the  figure  is  a  rectangle. 

Ex.  12.    Prove  that  a  parallelogram  inscribed  in  a  circle  is  a  rectangle. 


SUPPLEMENTARY  EXERCISES  285 

Ex.  13.     If  AB  and  AF  are  tangents  to  the  circle  whose  center  is  0 
and  E  is  any  point  in  major  arc  Z>jP,  then,  Z  DEF  =  90°  —  \Z.A. 
Suggestions.  —  1.    Draw  OD  and  OF. 
2.    Compare  LE  \\\ih.  LDOF,  and  ADOF  with  LA. 

Ex.  14.  If  AB  and  AC  are  tangents  to  a  circle  whose  center  is  O 
from  a  point  A,  touching  the  circle  at  B  and  C  respectively,  and  D  is  any 
point  on  the  minor  arc  BC,  then  ZBDC  =  90''  -h^ZA.    (Ex.  71,  p.  117.) 

Ex.  15.  ABCD  is  a  quadrilateral  inscribed  in  a  circle.  Another 
circle  is  drawn  upon  AB  as  chord,  meeting  AB  and  CB  at  JS^  and  F  re- 
spectively.    Prove  chords  BC  and  ^F  parallel.     (Ex.  71,  p.  117.) 

Ex.  16.  If  the  opposite  angles  of  a  quadrilateral  ABCB  are  supple- 
mentary, a  circle  can  be  circumscribed  about  the  quadrilateral. 

Suggestions.  —  1.  Assume  that  D  falls  outside  the  O  through  A,  B,  and 
C,  and  that  the  0  cuts  CD  at  E. 

2.  Derive  two  contradictory  facts  about  LD  and  Z.AEC,  using  the 
hypothesis  and  Ex.  71,  p.  117. 

3.  Next,  assume  that  D  falls  inside  the  0  ABC  and  complete  the  indirect 
proof. 

Ex.  17.  If  a  right  triangle  has  for  its  hypotenuse  the  side  of  a  square 
and  lies  outside  the  square,  the  straight  line  drawn  from  the  center  of  the 
square  to  the  vertex  of  the  right  angle  bisects  the  right  angle. 

Suggestio7i. —The  0  on  the  hypotenuse  as  diameter  must  pass  through 
the  center  of  the  square  and  also  through  the  vertex  of  the  right  angle. 
(Ex.  16,  p.  285.)    Draw  this  circle. 

Ex.  18.  The  perpendiculars  drawn  from  the  vertices  of  a  triangle  to 
the  opposite  sides  are  the  bisectors  of  the  angles  of  the  triangle  formed 
by  joining  the  feet  of  the  perpendiculars. 

Suggestions.  —  1.  ©  can  be  circumscribed  about 
quadrilaterals  BDOF,  CDOE,  and  AEOF.  (Ex. 
16,  p.  285.) 

2.  Compare  Z  ODF  with  L  OBF,  and  Z  ODE 
with  Z.OCE. 

3.  Compare  Z  OBF  with  Z  OCE,  by  connecting 
each  with  Z^^C 

4.  Then  AD  bisects  Z  EDF.       5.  Similarly  for  Z  DEF  and  Z  DFE. 

Ex.  19.  Construct  the  triangle  having  given  the  feet  of  the  perpen- 
diculars from  the  vertices  to  the  opposite  sides.     (Recall  Ex.  18,  p.  286.) 

Ex.  20.  If  sides  AB  and  BC  of  inscribed  hexagon  ABCDEF  are 
parallel  to  sides  2)^  and  EF  respectively,  prove  side  AF  parallel  to  side 
CD. 


286  SUPPLEMENTARY  EXERCISES 

Ex.  21.  If  a  circle  be  drawn  upon  the  radius  of  another  circle  as 
diameter,  any  chord  of  the  greater  circle  passing  through  the  point  of 
contact  of  the  circles  is  bisected  by  the  smaller  circle. 

Suggestion.  —  Recall  §  148. 

Ex.  22.  Prove  Prop.  XXI,  Book  II,  by  drawing  through  B  a  chord 
parallel  to  CD.     (Recall  §  208. ) 

Ex.  23.  If  sides  AB  and  BC  of  inscribed  quadrilateral  ABCD 
subtend  arcs  of  69''  and  112°  respectively,  and  Z  AED  between  the 
diagonals  is  87°,  how  many  degrees  are  there  in  each  angle  of  the 
quadrilateral  ? 

Suggestions.  —  1.  Let  x  =  AD  and  y  =  I)C.  Determine  these  arcs  alge- 
braically. 

2.    Then  determine  the  size  of  each  of  the  required  angles. 

Ex.  24.  Prove  Prop.  XXII,  Book  II,  by  drawing  through  B  a  chord 
parallel  to  (7i>.     (Recall  §208.) 

Ex.  25.  Prove  that  the  measure  of  the  angle  between  two  tangents 
is  the  supplement  of  the  measure  of  the  smaller  of  the  two  intercepted 
arcs. 

Suggestion.  —  After  obtaining  the  measure  of  the  angle,  substitute  in  it 
for  the  larger  arc  the  value  of  that  arc  in  terms  of  the  smaller  arc. 

Ex.  26.  If  sides  AB.,  BC,  and  CD  of  an  inscribed  quadrilateral  sub- 
tend arcs  of  99°,  106°,  and  78°  respectively,  and  sides  BA  and  CD 
extended  meet  at  E^  and  sides  AD  and  5(7  at  F.,  find  the  number  of 
degrees  in  Z  AED  and  Z  AFB. 

Ex.  27.  li  A  A.,  B,  and  C  of  circumscribed  quadrilateral  ABCD  are 
128°,  67°,  and  112°,  respectively,  and  sides  AB,  BC,  CD,  and  DA  are 
tangent  to  the  circle  at  points  E,  F,  G,  and  H  respectively,  find  the 
number  of  degrees  in  each  angle  of  the  quadrilateral  EFGH. 

Ex.  28.  If  AB  and  AC  are  the  tangents  to  a  circle  from  a  point  A, 
and  D  is  any  point  on  the  major  arc  subtended  by  chord  BC,  prove  that 
Z  ABD  +  Z  ^  CD  is  constant. 

Suggestions.  —  !.    AABD -{- AACD  =  360°  -  /lA  -  /.D.    Why? 

2.    Substitute  for  Z.A  and  Z.D  their  measures. 

Ex.  29.  If  ABCD  is  a  circumscribed  quadrilateral,  prove  that  the 
angle  between  the  lines  joining  the  opposite  points  of  contact  equals 
\{/.A-\-  Z.C)  or  is  supplementary  to  it. 

Suggestion.  —  Find  the  measure  of  each  of  the  angles.  Add  the  measure 
of /^and  ZC 


SUPPLEMENTARY   EXERCISES  287 

Ex.  30.     ABCD  is  a  quadrilateral  inscribed 
in  a  circle.     If  sides  AB  and   DC   extended 
intersect  at  JF,  and  AD  and  BC  extended  inter- 
sect at  F,  prove  that  the  bisectors  of  Z  E  and    H^ 
Z  F  are  perpendicular. 

Suggestions.  — 1.    AM+ AH+ KC  ■\-(jL  must    aN 
=  180°. 

2.    ^3/  enters  in  the  measure  of  Z  AEM,  and 
u4^in  that  of  LAFH.    Express  these  measures  and  add  the  results. 

This  will  give  a  start  on  the  proof. 

Construct  the  A  ABC  having  given  : 


Ex. 

31. 

a,     K,     he. 

Ex. 

32. 

a,     he,    to. 

Ex. 

33. 

A,     C,     to. 

Ex. 

34. 

c,     he,     nic' 

Ex. 

39. 

Construct  a 

Ex. 

35. 

a, 

&, 

A. 

Ex. 

36. 

A, 

B, 

he. 

Ex. 

37. 

b, 

ma, 

C. 

Ex. 

38. 

A, 

tA, 

ha. 

Construct  a  right  triangle  having  given  the  altitude  upon 
the  hypotenuse  and  one  of  the  legs  of  the  triangle. 

Ex.  40.     Construct  a  right  triangle  having  given  the  altitude  upon  the 
hypotenuse  and  one  of  the  acute  angles. 

/^ 
Ex.  41.     Construct  a  triangle    having  given  the  /  Nv 

mid-points  of  its  sides.  ^;" /v 

Suggestion.  —Kow  does  C'B'  compare  with  BC?  /       \/        \^ 

Ex.  42.     Construct  a  tangent  to  an  arc  of  a  circle  at  a  given  point  of 
the  arc  without  using  the  center  of  the  circle. 

Ex.  43.    Construct  a  tangent  to  a  given  circle  which  will  be  perpen- 
dicular to  a  given  straight  line. 

Ex.  44.    Given  the  mid-point  of  a  chord  of  a  circle,  construct  the 
chord. 

Ex.  45.  Construct  a  parallel  to  side  BC  of 
A  ABC  meeting  AB  and  ^C  at  D  and  E  respec- 
tively, so  that  DE  will  equal  the  sum  of  BD  and  CE. 

Ex.  46.  Inscribe  a  square  within  a  given  right 
triangle  having  one  of  its  angles  coincident  with  the 
right  angle  of  the  triangle  and  the  opposite  vertex  lying  on  the  hypotenuse. 

Suggestion.  —  In  the  analysis  figure,  draw  the  diagonal  from  the  vertex  of 
the  right  triangle. 


288 


SUPPLEMENTARY  EXERCISES 


Ex.  47.  Construct  a  rhombus  within  a  given  tri- 
angle, having  one  angle  coincident  with  an  angle  of  the 
triangle,  and  the  opposite  vertex  lying  on  the  opposite 
side  of  the  triangle. 

Ex.  48.  Construct  a  square  which  will  have  its  ver- 
tices on  the  sides  of  a  given  rhombus. 

Suggestion.  — Make  an  analysis  based  upon  the  adjoining 
figure. 


Ex.  49.     Construct  two  tangents  to  a  given  circle  which  will  make  a 
given  angle  with  the  circle. 

Suggestion. — Draw  the  given  angle  at  the  center  of  the  circle. 


Ex.  50.     Given  an  angle  of  a  triangle  and 

the  segments  of  the  opposite  side  made  by  the 

altitude  drawn  to  that  side.  Construct  the  tri- 
angle. 

Ex.  51.  Construct  a  A  ABC  having  given 
c,  6,  and  Wo.  Make  an  analysis  based  upon  the 
adjoining  figure. 


Ex.  52.  Through  a  given  point  outside  a  circle,  construct  a  secant 
whose  internal  and  external  segments  will  be  equal. 

Suggestion.  —  For  the  analysis  figure,  connect  the  center  of  the  circle  with 
the  given  point,  and  also  with  the  points  of  intersection  of  the  secant  and  the 
circle.    Recall  Ex.  51,  p.  288. 


Ex.  53.  Given  the  base,  an  adjacent  acute  angle, 
and  the  difference  between  the  other  two  sides  of  the 
triangle,  construct  the  triangle. 

Suggestions.  — 1.  LetCD=CB.    Then AD=AC-BC. 
2.    Then  A  ABB  can  be  made  the  basis  of  the  con- 
struction. 


Ex.  54.  Given  the  base  of  a  triangle,  an  adjacent 
angle,  and  the  sum  of  the  other  two  sides,  construct 
the  triangle. 

Suggestions.  — 1.    Let  AD  =  AC+  CB.    Draw  CB  1 


BD. 


2.    A  ABB  can  be  made  the  basis  of  the  construction. 


/^' 


SUPPLEMENTARY  EXERCISES  289 


Ex.  55.  Construct  full  size  the  pattern  for  the 
faces  of  a  mission  lamp  as  shown  in  the  adjoining 
figure,  using  the  dimensions  indicated. 


om 


Ex.  56.  Construct  a  circle  tangent  to  a  given  line  and  having  its 
center  at  a  given  point  not  on  the  line. 

Ex.  57.  Construct  a  circle  which  will  be  tangent  to  each  of  two 
parallels  and  will  pass  through  a  given  point  lying  between  the  parallels. 

Ex.  58.  Construct  a  circle  having  its  center  in  a  given  line,  and  pass- 
ing through  two  points  not  in  the  line. 

Ex.  59.  Construct  a  circle  with  given  radius  which  will  be  tangent 
to  a  given  circle  and  will  pass  through  a  given  point  outside  of  the  circle. 

Ex.  60.  Construct  a  circle  with  given  radius  which  will  be  tangent  to 
a  given  circle  and  pass  through  a  given  point  inside  of  the  circle. 

Ex.  61.  Construct  a  circle  with  a  given  radius  which  will  be  tangent 
to  a  given  line  and  also  to  a  given  circle. 

Ex.  62.  Construct  a  circle  which  will  be  tangent  to  a  given  circle  at 
a  given  point  on  it  and  also  tangent  to  a  given  straight  line. 

Ex.  63.  Construct  a  circle  which  will  be  tangent  to  a  given  circle  at 
a  given  point  on  it  and  also  pass  through  a  given  point  outside  of  the  circle. 

BOOK  III 

Ex.  1.  Prove  the  theorem  of  §  268  on  the  hypothesis  that  AD  :  DB 
=  AE :  EC. 

Suggestion.  —  Use  the  same  construction.  Write  the  hypothesis  by  com- 
position, and  use  §  262. 

Ex.  2.  Let  P  be  any  point  not  in  line  AB  and  B  any  point  in  AB. 
Let  ^  be  a  point  in  segment  PB,  such  that  FS  :  PB  =  1:3.  Suppose 
that  B  moves  along  AB.     What  is  the  locus  of  point  S  ? 

Ex.  3.  XY  is  parallel  to  the  side  AB  of  A  OAB,  meeting  OA  at  X 
and  OB  at  Y.  Point  C  is  taken  between  X  and  A  of  OA,  and  ^C  is 
drawn.     XZ  is  drawn  parallel  to  BC,  meeting  YB  at  Z.    Prove  CY\\  AZ. 

Suggestion.  —  Try  to  prove  OC  •  OZ  =  OA  ■  OY;  then  use  §  252. 

Ex.  4.  State  and  prove  the  converse  of  Prop.  IV,  §  270.  (Fi§.  of 
Prop.  IV.    Prove  Z  BAD=Z  CAD.   Extend  CA  to  E^  making  AE=AB.) 


290  SUPPLEMENTARY  EXERCISES 

Ex.  5.  The  sides  of  a  triangle  are  a,  &,  and  c,  respectively.  Derive 
formulae  for  the  segments  of  side  c  made  by  the  bisector  of  Z  G. 

Ex.  6.  AB  is  the  hypotenuse  of  right  A  ABC.  If  perpendiculars  be 
drawn  to  AB  at  A  and  B^  meeting  AC  extended  at  2>,  and  BG  extended 
at  jE',  prove  ^  ACE  and  BCD  similar. 

Ex.  7.  If  altitudes  AB  and  GE  of  A  ABC  intersect  at  F,  prove 
AF:AB  =  EF.BD. 

Ex.  8.     ^^  is  a  chord  of  a  circle,  and  CE  is  any  chord  drawn  through 
the  middle  point  C  of  arc  AB,  cutting  chord  ^^  at  D. 
Prove  ^O  is  a  mean  proportional  between  CD  and  CE. 

Ex.  9.  Two  circles  are  tangent  internally  at  C.  GA  is  drawn  meet- 
ing the  smaller  circle  at  B  and  the  larger  at  A  ;  CE  is  drawn  meeting  the 
smaller  circle  at  Z>  and  the  larger  at  E.     Prove  GB :  GA  =  CD  :  GE. 

Ex.  10.  The  diagonals  of  a  trapezoid,  whose  bases  are  AD  and  BG, 
intersect  at  E.    If  AE  =9,  EC=  3,  and  BB  =  16,  find  BE  and  ED. 

Suggestion.  — Fvoye  AE  :  EC=  DE  :  EB. 

Ex.  11.  Let  ^C  be  the  hypotenuse  of  right  A  ABC,  and  E  and  F  be 
any  points  on  AB  and  BG  respectively  ;  let  JE'i>  and  FO  be  perpendic- 
ulars to  AC,  meeting  ylC  at  Z>  and  G  respectively.  Prove  AE :  FC 
=  ED  :  GC.     (Recall  §  109.) 

Ex.  12.  Z  ^  of  A  ABC  is  a  right  angle.  DEFG  is  a  square  having  E 
and  F on  BG,  D  on  AC,  and  G  on  AB.     Prove  GE:  EF=  EF :  FB. 

Suggestion.  — Compare  A  (7i)jE;  and  A  BFG. 

Ex.  13.  AB  and  ^C  are  the  tangents  to  a  circle  0  from  point  A.  If 
CD  is  drawn  perpendicular  to  OB  produced  at  D,  then  AB  :  OB  =  BD  : 
CD. 

Suggestion. — Draw  OA  and  BG.    Prove  OA  1  BG. 

Ex.  14.  AABCis^Lu  isosceles  triangle.  If  the  perpendicular  to  AB 
at  A  meets  base  BG,  extended  if  necessary,  at  E,  and  D  is  the  mid-point 
of  BE,  then  AB  is  the  mean  proportional  between  BG  and  BD. 

Suggestion.  —  Recall  §284  and  Ex.  175,  Book  I. 

Ex.  15.  Let  r  be  the  radius  of  a  circle  and  c  be  the  distance  from  the 
center  of  the  circle  to  a  point  P  outside  the  circle.  Express  the  length  of 
the  tangent  to  the  circle  from  P,  in  terms  of  r  and  c. 

Ex.  16.  What  is  the  length  of  the  tangent  to  a  circle  whose  diameter 
is  16,  from  a  point  whose  distance  from  the  center  is  17  ? 


SUPPLEMENTARY   EXERCISES  291 

Ex.  17.  Prove  that  the  tangents  to  two  inter- 
secting circles  from  any  point  in  their  common 
chord  produced  are  equal.     (Figure  adjoining.) 

Ex.  18.  If  two  circles  intersect,  their  common 
chord  produced  bisects  their  common  tangents. 

Ex.  19.     If  the  altitude  be  drawn  to  the  hypot- 
enuse of  a  right  triangle,  the  segments  of  the  hypotenuse  have  the  same 
ratio  as  the  squares  of  the  adjacent  legs. 

Ex.  20.  What  is  the  length  of  a  chord  of  a  circle  which  is  6  in.  from 
the  center,  if  the  radius  is  10  in.  ? 

Ex.  21.    The  equal  angles  of  an  isosceles  triangle  are  each  30°,  and  the 
equal  sides  are  each  8  in.  in  length.     What  is  the  length  of  the  base  ? 
Suggestion.  —  Recall  Ex.  128,  Book  I. 

Ex.  22.  Find  the  altitude  to  the  base  of  an  isosceles  triangle  if  the 
base  is  8  inches  and  the  sides  are  each  10  inches  in  length. 

Ex.  23.  If  the  equal  sides  of  an  isosceles  right  triangle  are  each  18 
in.  in  length,  what  is  the  length  of  the  median  drawn  from  the  vertex  of 
the  right  angle  ? 

Ex.  24.  One  of  the  non-parallel  sides  of  a  trapezoid  is  perpendicular 
to  the  bases.  If  the  length  of  this  side  is  40,  and  of  the  parallel  sides  31 
and  22,  respectively,  what  is  the  length  of  the  other  side  ? 

Ex.  25.  If  the  length  of  the  common  chord  of  two  intersecting  circles 
is  16,  and  their  radii  are  10  and  17,  what  is  the  distance  between  their 
centera  ? 

Ex.  26.  If  BC  is  the  hypotenuse  of  right  triangle  ABC,  prove 
(a  +  6-|-c)2  =  2(a  +  c){a  +  b). 

Ex.  27.  If  the  diagonals  of  a  rhombus  are  m  and  n  respectively, 
derive  a  formula  for  the  perimeter  of  the  rhombus. 

Ex.  28.  The  diameter  which  bisects  a  chord  12  in.  long  is  20  in.  in 
length.  Find  the  distance  from  either  extremity  of  the  chord  to  the 
extremities  of  the  diameter. 

Suggestions.  —  1.  Let  x  represent  one  segment  of  the  diameter  made  by  the 
chord.  2.   Recall  §  289. 

Ex.  29.  The  radius  of  a  circle  is  16  in.  Find  the  length  of  the  chord 
which  joins  the  points  of  contact  of  two  tangents,  each  30  in.  in  length, 
drawn  to  the  circle  from  a  point  outside  the  circle. 

Suggestions.  —  1.  Draw  the  radii  to  the  points  of  contact.    2.  Recall  §  288. 


292  SUPPLEMENTARY  EXERCISES 

Ex.  30.  Two  parallel  chords  on  opposite  sides  of  the  center  of  a  circle 
are  48  in.  and  14  in.  long,  respectively,  and  the  distance  between  their 
mid-points  is  31  in.     What  is  the  diameter  of  the  circle  ? 

Suggestion.  —  Let  x  represent  the  distance  from  the  center  to  the  middle 
point  of  one  chord,  and  31  —  x  the  distance  from  the  center  to  the  middle  point 
of  the  other.  Then  the  square  of  the  radius  may  be  expressed  in  two  ways  in 
terms  of  x. 

Ex.  31.  The  parallel  sides,  AD  and  BC,  of  a  circumscribed  isosceles 
trapezoid  are  18  and  6  respectively.     Find  the  diameter  of  the  circle. 

Suggestions.  —  1.  Recall  Ex.  35,  Book  II. 

2.  Through  B,  draw  BE  II  CD,  meeting  AD  at  E. 

Ex.  32.  The  diameters  of  two  circles  are  12  and  28,  respectively,  and 
the  distance  between  their  centers  is  29.  Find  the  length  of  the  common 
internal  tangent. 

Suggestion.  —  Find  the  1  drawn  from  the  center  of  the  smaller  0  to  the 
radius  of  the  greater  0  extended  through  the  point  of  contact. 

Ex.  33.  Prove  that  the  square  of  the  common  tangent  to  two  circles 
which  are  tangent  to  each  other  externally  is  equal  to  4  times  the  product 
of  their  radii. 

Ex.  34.  If  D  is  the  mid-point  of  leg  BC  of  right  triangle  ABC,  prove 
that  the  square  of  the  hypotenuse  AB  exceeds  3  times  the  square  of  CD 
by  the  square  of  AD. 

Ex.  35.  If  AB  is  the  base  of  isosceles  triangle  ABC  and  AD  is 
perpendicular  to  BC,  prove  AB^  +  BC^  +  AC^  =  3  AD^  +  2  CD^  +  BD^ 

Ex.  36.  If  D  is  the  mid-point  of  leg  BC  of  right  triangle  ABC,  and 
DE  is  drawn  perpendicular  to  hypotenuse  AB,  prove  AE   —  BE    =  ACT. 

Ex.  37.  If  in  right  triangle  ABC,  acute  angle  B  is  double  acute 
angle  A,  prove  AC^  =  S  BC^. 

Suggestion.  —  Recall  Ex.  128,  Book  I. 

Ex.  38.  Prove  that  the  sum  of  the  squares  of  the 
distances  of  any  point  on  a  circle  from  the  vertices  of 
an  inscribed  square  is  equal  to  twice  the  square  of  the 
diameter  of  the  circle. 

Ex.  39.  If  ABC  and  ADC  are  angles  inscribed 
in  a  semicircle,  and  AE  and  CF  are  drawn  perpen- 
dicular to  BD  extended,  prove  E 

BE^  +  BF'^  =  DE^  +  DF\ 


SUPPLEMENTARY  EXERCISES  293 


Ex.  40.     If  lines  be  drawn  from  any  point  P  to 
the  vertices  of  rectangle  ABCD^  prove  that 

PA^  +  PC^  =  P&  +  Plf. 

A 

t 

Ex.  41.     Inscribe  in  a  given  circle  a  triangle  similar  to  a  given  triangle. 

Suggestion.  —  Circumscribe  about  the  given  A  a  0,  and  draw  radii  to  the 
vertices.    Recall  §  293. 

Ex.  42.  Construct  a  right  triangle  having  given  its  perimeter  and  an 
acute  angle. 

Suggestion.  —  Any  right  triangle  containing  the  given  acute  angle  will  be 
similar  to  the  required  triangle.  The  sides  of  the  required  triangle  can  be  de- 
termined by  §  297. 

Ex.  43.  The  perimeter  of  one  of  two  similar  polygons  is  153  in.  ; 
the  shortest  side  of  this  polygon  is  18  in.  The  shortest  side  of  a  similar 
polygon  is  24  in.  ;  what  is  the  perimeter  of  the  second  polygon  ? 

Ex.  44.     The  adjoining  figure  is  similar  to  the        „ 

boundary  of  an  irregular  field  of  a  farm  ;  the  ratio 
of  similitude  of  the  figure  and  the  boundary  of  the 
field  is  1  :  2400.  Determine  the  perimeter  of  the 
field  itself  by  first  finding  the  perimeter  of  the  ad- 
joining figure  and  then  applying  §  297. 

Ex.  45.  If  JE7  is  the  mid-point  of  one  of  the  parallel  sides  BC,  of 
trapezoid  ABCD,  and  AE  and  DE  extended  meet  DC  and  AB  extended 
at  i^and  G  respectively,  then  FO  is  parallel  to  BC. 

Suggestion.  —  GF II  BC  if  GB  :GA  =  FE:  FA. 

Ex.  46.     A  ABC  and  yl'i?C  have  their  vertices     KT"  ----^A' 

A  and  A'  in  a  line  parallel  to  their  common  base      d\  ^v — y'     /^' 
BC.     If  a  parallel  io  BC  cuts  AB  at  i)  and  ^O  at        \   ^X^    / 
E,  A'B  at  D'  and  ^'C  at  E\  then  DE  =  D'E'.  q^ — -^ 

Suggestion.  —  Prove  DE  :  BC  =  D'E' :  BC. 

Ex.  47.  If  AB  and  CD  are  equal  and  parallel  segments,  prove  that 
p^^  equals  p^^,  where  m  is  any  line. 

Ex.  48.   If  AD  and  BE  are  the  perpendiculars  from  vertices  A  and  B, 
respectively,  of  acute-angled  triangle  ABC  to  the  opposite  sides,  prove 
ACxAE  +  BCxBD  =  AB\ 

Suggestion.  —  Find  2ACx  AE  by  §  310,  and  in  like  manner  find  2BCx  BD 
Then  add. 


294  SUPPLEMENTARY   EXERCISES 

Ex.  49.     In  triangle  ABC,  if  angle  C  equals  120°,  prove 

AB^^  =  BC^  +  AC^  +  BCx  AC. 
Suggestion.  —  Recall  §  311. 

Bx.  50.    If  a  line  be  drawn  from  vertex  C  of  isosceles  triangle  ABC, 
meeting  base  AB  extended  at  D,  prove  CB^  —  CB^  =  AD  x  BD. 
Suggestion.  — Apply  §  311  in  A  BCD. 

Ex.  51.  From  the  conclusion  of  §  311,  derive  a  formula  for  pj  in  terms 
of  a,  6,  and  c. 

Ex.  52.    In  any  triangle,  the  product  of  any  two 

sides  is  equal  to  the  product  of  the  segments  of  the  F>^         '-'P^'^'n 

third  side  formed  by  the  bisector  of  the.  exterior  ^^^>fC;        \   '^' 

angle  at  the  opposite  vertex,  minus  the  square  of  /^'^J\\  \  ; 

the  bisector.  q  ^ X — — 4'q 

Prove       AB  X  AC  =  DB  X  DC  -  AD\  ^  "~ " 

Suggestions.  —  1.  The  solution  is  similar  to  that  of  §  318. 

2.   Firstprove  A^£Z)~  A^C^. 

Ex.  53.  DEFG  is  a  square  having  its  vertices  D  and  E  on  sides  AB 
and  BC  respectively  of  triangle  ABC  and  its  vertices  F  and  O  on  side  AC. 
Let  BH  be  II  to  AC,  meeting  AE  extended  at  G  ;  let  UK  he  ±  AC  and 
BT±AC.     Prove  BHKT  is  a  square, 

Ex.  54.  In  a  given  triangle,  construct  a  square  which  shall  have  two 
vertices  lying  on  one  side  of  the  triangle  and  having  its  other  two  vertices 
on  the  other  two  sides  of  the  triangle,  one  on  each  side. 

Ex.  55.  Construct  a  square  which  will  have  two  of  its  vertices  on  a 
diameter  of  a  given  circle,  and  the  remaining  two  vertices  on  the  semicircle 
constructed  on  this  diameter. 

Ex.  56.  Circumscribe  about  a  given  circle  a  triangle  similar  to  a  given 
triangle. 

Suggestion.  —  Inscribe  in  the  given  triangle  a  circle  and  draw  radii  to  the 
points  of  tangency. 

BOOK  IV 

Ex.  1.  The  sides  of  a  triangular  field  are  10  rd.,  8  rd.,  and  9  rd. 
respectively.  Make  a  scale  drawing  of  the  boundary  of  the  field  on  coor- 
dinate paper,  and  estimate  the  area  of  the  field. 

Ex.  2.  Angle  B  of  A  ABC  is  a  right  Z.  D  and  E  are  the  mid-points 
of  AB  and  AC  respectively.  CF,  perpendicular  to  BC  at  C,  meets  DE 
extended  at  F.     Prove  A  J-jSC  =  □  BCFD. 


SUPPLEMENTARY  EXERCISES  295 

Ex.  3.  E  is  any  point  on  diagonal  ^C  of 
O  ABCD.  Tlirough  E,  parallels  to  AD  and 
AB  are  drawn,  meeting  AB  and  CD  at  F  and 
irrespectively,  and  BC  and  AD  at  G  and  K  re- 
spectively. 

Prove  CJFBGE  =  CD  EHDK. 

Ex.  4.  All  the  lots  of  a  certain  city  block  are  rectangular  and  125  ft. 
in  depth  (from  front  to  back).  Compare  two  lots  A  and  B  if  the  frontage 
of  Lot  A  is  40  ft.  and  that  of  Lot  J5  is  60  ft.     (Do  not  obtain  their  areas.) 

Ex.  5.     Two  rectangles  B\  and  B<2  have  equal  altitudes. 

(a)  What  part  of  B-z  is  B\  if  the  base  of  Bi  is  5  and  the  base  of  7^2  is  8  ? 

(6)  What  is  the  ratio  of  ^i  to  B2  if  the  bases  are  25  and  10  respec- 
tively ? 

Ex.  6.  Divide  a  given  triangle  into  three  equal  parts  by  lines  drawn 
through  one  of  its  vertices. 

Ex.  7.  Determine  the  area  of  the  triangle  whose  sides  are  25,  17, 
and  28. 

Ex.  8.  If  h  is  the  base  and  s  is  one  of  the  equal  sides  of  an  isosceles 
triangle,  prove  that  the  area  is  ^  6  V4  s^  —  ft^. 

Ex.  9.  The  area  of  an  isosceles  right  triangle  is  81  sq.  in.  Deter- 
mine its  hypotenuse. 

Suggestion.  —  I^iet  x  represent  one  of  the  sides.  Determine  x  and  then  de- 
termiue  the  hypotenuse. 

Ex.  10.  The  area  of  an  equilateral  triangle  is  9V3.  Determine  its 
side. 

Suggestion.  —  Use  the  formula  proved  in  Ex.  29,  Book  IV. 

Ex.  11.  The  altitude  of  an  equilateral  triangle  is  3.  Determine  its 
area. 

Suggestion.  — Let  x  represent  one  side.  Determine  x  and  then  determine 
the  area. 

Ex.  12.  The  area  of  an  equilateral  triangle  is  16  V3.  Determine  its 
altitude. 

Ex.  13.  The  area  of  a  rhombus  is  240  sq.  in.  and  its  side  is  17  in. 
Find  its  diagonals. 

Suggestion. —  Represent  the  diagonals  by  2  a;  and  2  y.    Proceed  algebraically. 

Ex.  14.  One  diagonal  of  a  rhombus  is  five  thirds  the  other;  the  dif- 
ference of  the  diagonals  is  8  in.     Determine  the  area  of  the  rhombus. 

Ex.  15.  The  segments  of  the  hypotenuse  of  a  right  triangle  made  by 
the  altitude  drawn  to  the  hypotenuse  are  5^  and  9f  respectively.  Deter- 
mine the  area  of  the  triangle.     (§  288.) 


296 


SUPPLEMENTARY   EXERCISES 


Ex.  16.     The  sides  of  A  ABO  are  AB  =  13,  BC  =  14,  and  AC  =  15. 
Bisector  AD  ot  Z  A  meets  BG  ?itD.     Find  tlie  areas  of  ^  ABD  and  ACD. 

Suggestion.  —  1.  Compute  the  altitude  Aa.     (§313.) 
2.  Determine  BD  and  i>(7  by  §  270. 

Ex.  17.     If  Z>  and  E  are  the  mid-points  of  sides  BC  and  AC  respec- 
tively of  A^^O,  prove  A  ABD  =.  A  ABE. 

Suggestion.  — Compare  the  altitudes  to  AB  from  D  and  E,   ' 

Ex.  18.     If  diagonal  AC  of  quadrilateral  ABCD  bisects  diagonal  BD^ 
thenAJJ5a  =  A^D(7. 

Ex.  19.  Two  equal  triangles  have  a  common 
base,  and  lie  on  opposite  sides  of  it.  Prove  that 
the  base,  extended  if  necessary,  bisects  the  line 
joining  their  vertices.     (Prove  CD  =  CD.) 

Ex-  20.  If  EF  is  any  straight  line  drawn 
through  the  point  of  intersection  of  the  diagonals 
oi  O  ABCD,  meeting  sides  AD  and  BC  at  E 
and  F  respectively,  then  A  BEE  =  A  CED. 

Suggestion.  —  Does  BF  =  ED  ? 

Ex.  21.  If  E,  F,  G,  and  ^  are  the  mid-points  of 
sides  AB,  BC,  CD,  and  DA,  respectively,  of  quad- 
rilateral ABCD,  prove  ^i?'(y/i"  a  parallelogram  equal 
to  one  half  ABCD.  A 

Ex.  22.  Prove  that  the  sum  of  the  perpendiculars 
from  any  point  within  an  equilateral  triangle  to  the 
three  sides  is  equal  to  the  altitude  of  the  triangle. 

Suggestions.  — -i.  A  J5PC+A  BPA-^A  APC=A  ABC. 

2.  Express  the  area  of  each  triangle  and  substitute  in 
this  equation. 

Ex.  23.     If  E  is  any  point  in  side  BCoi  CJ  ABCD,  and  DE  is  drawn, 
meeting  AB  extended  at  F,  prove  A  ABE  equals  A  CEF. 

Suggestion.— Comp2kTe  AFCD  with  O  ABCD. 


Ex.  24.  Prove  that  the  area  of  a  triangle  is  equal 
to  one  half  the  product  of  its  perimeter  by  the  radius 
of  the  inscribed  circle. 


De 


SUPPLEMENTARY  EXERCISES  297 

Ez.  25.  A  circle  whose  diameter  is  12  is  inscribed  in  a  quadrilateral 
whose  perimeter  is  50.    Find  the  area  of  the  quadrilateral. 

Ex.  26.  If  the  sides  of  a  triangle  are  15,  41,  and  52,  determine  the 
radius  of  the  inscribed  circle. 

Suggestions.  —  1.   Find  the  area  of  the  triangle. 
2.  Make  use  of  the  fact  proved  in  Ex.  24. 

Ex.  27.     If  Z>  is  the  mid-point  of  side  BC  of  A  ABC,  E  the  mid-point 
of  AD,  F  of  BE,  and  G  of  CF,  then  AABC  =  %A  EFG. 
Suggestion.  —  Draw  EC. 

Ex.  28.     If  BE  and  CF  are  medians  drawn  from  ^ 

vertices  B  and  C  of  A  ABC,  intersecting  at  D,  prove  /  Nv 

A  BCD  equals  quadrilateral  AEDF.  p/     ..\e 

Suggestion.  —  Compare  A   ABE  with  A  BEC  and  /^-''^^^D^^"«v^\ 

with  A  BFC.  B^=^— — - — -^^c 

Ex.  29.     Any  quadrilateral  ABCD  is  equivalent  to  ^^F 

a  triangle,  two  of  whose  sides  are  equal  to  diagonals  -'"'y    \ 

AC  and  BD,  respectively,  and  include  an  angle  equal  ^f\/^    '•  \ 

to  either  of  the  angles  between  AC  and  BD.  /  y^s.    \  /  1 

Prove    A  EFG  =  ABCD,    where    EF=AC,     and  J/        X\[    I 

EG  =  BD.  D\J^ 

G 
Suggestion.  — ComjpSiTe  A  DFG  with  A  BEF  and  then  with  A  ABC. 

Ex.  30.  Prove  that  two  triangles  are  equal  if  two  sides  of  one  equal 
respectively  two  sides  of  the  other  and  the  included  angles  are  supple- 
mentary. 

Suggestion.  —  Place  the  triangles  so  that  the  supplementary  angles  are 
adjacent  and  so  that  one  pair  of  equal  sides  coincide. 

Ex.  31.     On  coordinate  paper,  draw  the  pentagon  whose  vertices  are 
^  =  0,  0  ;  B  =  5,  0  ;   C  =  8,  3  ;  Z>  =  4,  9  ;  ^  =  0,  6. 

Determine  its  approximate  area  as  in  Ex.  1,  and  Ex.  2,  Book  IV. 
Construct  a  A  equal  to  the  pentagon.  Then  construct  its  base  and  alti- 
tude, and  compute  its  approximate  area. 

Ex.  32.  If,  in  the  figure  of  Prop.  X,  AB  =  9  in.,  A'B'  =  7  in.,  and 
the  area  of  A  A'B'C  is  147  sq.  in.,  find  the  area  of  A  ABC. 

Ex.  33.  The  area  of  a  certain  triangle  is  f  the  area  of  a  similar 
triangle.  If  the  altitude  of  the  first  is  4  ft.,  what  is  the  altitude  of  the 
second  ? 


298  SUPPLEMENTARY   EXERCISES 

Ex.  34.  If,  in  §343,  area  of  AA'B'C  =147  sq.  in.,  AB  =  9  in., 
and  A'B'  =  3  in.,  find  the  area  of  A  ABC. 

Ex.  35.  The  sides  AB  and  ^C  of  A  ABC  are  15  and  22,  respectively. 
From  a  point  D  in  AB^  a  parallel  to  BC  is  drawn  meeting  AC  a.t  E,  and 
dividing  the  triangle  into  two  equal  parts.     Find  AD  and  AE. 

Ex.  36.  If  similar  polygons  be  drawn  upon  the  legs  of  a  right  tri- 
angle as  homologous  sides,  the  polygon  drawn  upon  the  hypotenuse  is 
equal  to  the  sum  of  the  polygons  drawn  upon  the  legs. 

Suggestions.  —  1.  Compare  the  polygon  on  each  leg  with  the  one  on  the 
hypotenuse  by  §  344. 

2.    Add  the  resulting  equations  and  simplify. 

Ex.  37.  Construct  a  triangle  similar  to  two  given  similar  triangles 
and  equal  to  their  sum. 

Ex.  38.  Two  similar  triangles  have  homologous  sides  of  8  in.  and 
15  in.  respectively.  Find  the  homologous  side  of  a  similar  triangle  equal 
to  their  sura. 

Ex.  39.  Construct  a  triangle  similar  to  two  similar  triangles,  and 
equal  to  their  difference. 

Ex.  40.  If  the  area  of  a  polygon,  one  of  whose  sides  is  15  in.,  is  375 
sq.  in.,  what  is  the  area  of  a  similar  polygon  whose  homologous  side  is 
10  in.  ? 

Ex.  41.  If  the  area  of  a  polygon,  one  of  whose  sides  is  36  ft.,  is  648 
sq.  ft,,  what  is  the  homologous  side  of  a  similar  polygon  whose  area  is 
392  sq.  ft.  ? 

Ex.  42.  Construct  a  rectangle  having  a  given  altitude  and  equal  to 
a  given  parallelogram. 

Suggestion.  —  Recall  Ex.  78,  Book  IV. 

Ex.  43.  Construct  a  parallelogram  equal  to  a  given  parallelogram 
and  having  two  adjacent  sides  equal  to  given  segments  m  and  n  respec- 
tively. 

Ex.  44.  Construct  a  parallelogram  equal  to  a  given  parallelogram 
and  having  one  side  equal  to  a  given  segment  m,  and  one  diagonal  equal 
to  a  given  segment  n. 

Ex.  45.  Construct  a  right  triangle  equal  to  a  given  square,  having 
given  its  hypotenuse. 

Suggestion.  —  Determine  the  altitude  to  the  hypotenuse  as  in  Ex.  78,  Book 
IV  ;  then  construct  the  triangle,  using  the  methods  of  §  241. 

Ex.  46.  Construct  a  right  triangle  equal  to  a  given  triangle,  having 
given  its  hypotenuse. 


SUPPLEMENTARY   EXERCISES  299 


BOOK  V 

Ex.  1.  Prove  that  the  diagonals  drawn  from  one  vertex  of  a  regular 
polygon  having  n  vertices  to  each  of  the  other  vertices  divides  the  angle  at 
that  vertex  into  (n  —  2)  equal  parts. 

Ex.  2.  Prove  that  the  central  angle  of  any  regular  polygon  is  the 
supplement  of  the  vertex  angle  of  the  polygon. 

Ex.  3.  Prove  that  the  sum  of  the  perpendiculars  drawn  from  any 
point  within  a  regular  polygon  to  the  sides  of  the  polygon  is  equal  to  the 
apothera  multiplied  by  the  number  of  sides  of  the  polygon. 

Suggestions.  —  Connect  the  point  with  each  vertex.  Notice  that  the  sum 
of  the  triangles  so  formed  equals  the  polygon.  Express  the  area  of  each  tri- 
angle and  form  an  equation. 

Ex.  4.     In  the  figure  for  §  365  prove  that : 
(a)  S4  >  S8  >  «i6,  etc.     (See  §  362. ) 
(6)  a4  <  as  <  aie,  etc. 
(c)  ki  <  ^8  <  A:i6,  etc. 

Ex.  5.  Prove  that  an  equiangular  polygon  inscribed  in  a  circle  is  reg- 
ular if  the  number  of  sides  is  odd. 

Ex.  6.  Prove  that  an  equiangular  polygon  circumscribed  about  a  cir- 
cle is  regular. 

Suggestions.  —  1.  Draw  the  chords  joining  the  points  of  tangency. 
2.  Prove  the  resulting  As  : 

(a)   are  isosceles;  (6)  are  mutually  equiangular;  (c)  that  XY=  YZ,  etc. 
See  diagram  in  §  3G7. 
Complete  the  proof. 

Ex.  7.  Repeat  Ex.  14,  p.  227,  for  a  regular  octagon  circumscribed 
about  a  circle  of  radius  10. 

Ex.  8.  Prove  that  diagonal  AE  oi  regular  octagon  ABCDEFGH  is 
the  perpendicular-bisector  of  diagonal  BH. 

F 

Ex.  9.    In  the  adjoining  figure,  ABCD  and        /O'y^    '<    ^v\ 
EFOH  are  squares  inscribed  in  the  circle,  such        (/\     ^^\!y'     Y\\ 

that  AF=  FB  =  BO,  etc.     Is  BSTUVWXY  a  ^t\r"7,'^;"""'br 
regular  octagon  ?  \Nc'     ' 


H 


300 


SUPPLEMENTARY  EXERCISES 


Ex.  10.     Construct  a  Maltese  cross  having  the  dimen- 
sions indicated. 


Ex.  11.  Prove  that  the  construction  indicated  in 
the  adjoining  figure  serves  to  inscribe  a  regular  octa- 
gon in  the  square. 


Ex.  12.    A  regular  octagon  is  inscribed  in  a  circle  of  radius  10.    Com- 
pute S8,  i?8,  «8,  and  ^'8. 

Ex.  13.     Prove  that  for  a  regular  octagon  inscribed  in  a  circle  of 


radius  i?  : 

(a)  SB  =  -BV2-V2 


B 


(c)  a8=^V2+V2; 
2 


(&)  i)8  ==  8  i? V2  -  V2  ;  {d)  ki=2 i?2v'2. 

Ex.  14,     Construct  a  regular  octagon  having  its  sides  1  inch  long. 

Ex.  15.    What  is  the  relation  between  the  area  of  the  inscribed  and  of 
the  circumscribed  equilateral  triangles  of  a  given  circle  ? 

Ex.  16.    What  is  the  relation  between  the  perimeter  of  the  inscribed 
and  of  the  circumscribed  equilateral  triangles  of  a  given  circle  ? 

Ex.  17.   A  regular  hexagon  is  inscribed  in  a  circle  of  radius  r.    Prove  : 


(a)  s^  =  r)  (6)  ae  = 


rVS 


(c)  j96  =  6  r ;    (d)  Tc^  = 


3rV3 


2    '    ^  '  ^"  '    "  '      "  2 

Ex.  18.    A  regular  triangle  is  inscribed  in  a  circle  of  radius  r.     Prove 

3rVS;  (d)    kz 


(a)  S3  =  rVS  ;  (&)  as=-r;  (c)  ps 
2 


3r2V3. 


Ex.  19.     Prove  that  the  apothem  of   an   equilateral  triangle  is  one 
third  the  altitude  of  the  triangle. 

Ex.  20.     (a)  In  a  circle  of  radius  2.5  in.,  inscribe  a  regular  hexagon. 

(b)  Also  inscribe  in  the  same  circle  a  regular  triangle  and  a  regular 
12-gon. 

(c)  Prove  that  S3  >  se  >  S12,  etc. 

(d)  Prove  that  as  <  ae  <  «i2,  etc. 

(e)  Prove  thatps  <pe<pi2,  etc. 
(/)  Prove  that  ks  <  ke  <  ki2,  etc. 


SUPPLEMENTARY   EXERCISES  301 

Ex.  21.     What  is  the  perimeter  and  area  of  a  regular  circumscribed 
hexagon  about  a  circle  of  radius  10  ? 

Ex.  22.     Repeat  the  foregoing  exercise  for  a  circle  of  radius  B. 

Ex.  23.     What  is  the  perimeter  and  the  area  of  a  regular  triangle 
circumscribed  about  a  circle  of  radius  10  ? 

Ex.  24.     Repeat  the  foregoing  exercise  for  a  circle  of  radius  B. 

Ex.  25.  Prove  that  the  diagonals  AC^  BD^  CE^  etc.,  of  regular  hex- 
agon ABCDEFiorm  another  regular  hexagon. 

Suggestion.  — ^^Prove  that  a  circle  can  be  inscribed  in  the  inner  hexagon. 

Ex.  26.     Prove  that  the   area  of  the   inner  B^^;^ — ^ 

hexagon  of  the  foregoing  exercise  is  one  third  the 
area  of  ABC  DBF. 

Suggestion.  —  Express  the  area  of  each  polygon 
in  terras  of  the  radius  OB  of  ABC  DBF. 

Ex.  27.  Prove  that  the  area  of  a  regular  in- 
scribed hexagon  is  a  mean  proportional  between 
the  areas  of  an  inscribed  and  of  a  circumscribed  equilateral  triangle/ 

Suggestion.  —  Express  the  areas  of  each  in  terms  of  the  radius. 

Ex.  28.  In  a  given  equilateral  triangle,  inscribe  a  regular  hexagon 
having  two  of  its  vertices  lying  on  each  side  of  the  triangle. 

Ex.  29.  Construct  a  regular  hexagon  having  given  one  of  the  diag- 
onals joining  two  alternate  vertices. 

Ex.  30.  A  square  is  inscribed  in  an  equilateral  triangle  whose  side 
is  a,  having  two  vertices  in  one  side  of  the  triangle,  and  one  in  each  of 
the  other  sides.    Compute  the  area  of  the  square. 

Ex.  31.     A  regular  12-gon  is  inscribed  in  a  circle  of  radius  B.    Prove  : 
(a)  Si2  =  i?  V2  -  V3  ;  (c)  pi^  =  12  i?  V2  -  Vs  ; 

(6)  ai2  =  ^V2Wl;  id)k,,  =  SB^. 

Z 

Ex.  32.  If  the  diagonals  AC  and  BE  of  regular  pentagon  ABCDE 
intersect  at  F,  prove  that  BE  =  AE -\-  EF. 

Ex.  33.   Prove  that  the  figure  FOHKL  formed  by  parts  of  the  diag- 
onals of  regular  inscribed  petagon  ABCDE  is  also  a  regular  pentagon. 
(See  figure  on  page  302.) 


302  SUPPLEMENTARY  EXERCISES 

Suggestions. — Prove  that  a  circle  can  be  inscribed 
in  FGHKL. 

Ex.  34.  — In  the  figure  of  Prop.  VIII,  Book  V,     f"^   ^ 
prove  that  OM  is  the  side  of  a  regular  pentagon  in-        \  l 
scribed  in  the  circle  which  can  be  circumscribed  about 
/\OBM. 

Suggestion.  —  How  large  is  Z  OBM  ? 

Ex.  35.     Construct  a  regular  pentagon  having  given  one  of  its  sides. 

Ex.  36.  Construct  a  regular  pentagon  having  given  one  of  its  diago- 
nals. 

Ex.  37.  If  B  represents  the  radius  of  the  circle  circumscribed  about 
a  regular  decagon,  prove  : 

(a)  sio=_(V5-l);  (c)  pio  =  5i?(\/5  -  1)  ; 

(6)  aio=  — Vl0  +  2\/5;  (d)  A^io  =  ^^  VlO  -  2  V5. 

4  4 

Ex.  38.  Find  the  area  of  the  circle  inscribed  in  a  square  whose  area 
is  25. 

Ex.  39.  If  the  radius  of  a  circle  is  SVS,  what  is  the  area  of  the  sector 
whose  central  angle  is  150°  ? 

Ex.  40.  Find  the  radius  of  the  circle  equal  to  a  square  whose  side  is 
10. 

Ex.  41.  Find  the  radius  of  the  circle  whose  area  is  one  half  the  area 
of  the  circle  whose  radius  is  15. 

Ex.  42.  Find  the  area  of  the  square  inscribed  in  the  circle  whose  area 
is  196  IT  sq.  in. 

Ex.  43.  The  area  of  one  circle  is  ^  the  area  of  another.  Find  the 
radius  of  the  second  if  the  area  of  the  first  is  15. 

Ex.  44.  The  side  of  a  square  is  8.  Find  the  circumference  of  its  in- 
scribed and  circumscribed  circles. 

Ex.  45.  The  side  of  an  equilateral  triangle  is  6.  Find  the  area  of  its 
inscribed  and  circumscribed  circles. 

Ex.  46.  The  area  of  a  regular  hexagon  inscribed  in  a  circle  is  24  VS. 
What  is  the  area  of  the  circle  ? 

Ex.  47.  If  the  apothem  of  a  regular  hexagon  is  6,  what  is  the  area  of 
its  circumscribed  circle  ? 


SUPPLEMENTARY  EXERCISES  303 

Ez.  48.  Two  plots  of  ground,  one  a  square  and  one  a  circle,  each 
contain  70686  sq.  ft.  How  much  greater  is  the  perimeter  of  the  square 
than  the  length  of  the  circle  ? 

Ex.  49.  The  perimeter  of  a  regular  hexagon  circumscribed  about  a 
circle  is  12  VS.     What  is  the  circumference  of  the  circle  ? 

Ex.  50.     The  length  of  the  arc  subtended  by  the  side  of  a  regular  in- 
scribed 12-gon  18 -IT  in.     What  is  the  area  of  the  circle  ? 
°  3 

Ex.  51.  If  the  length  of  a  quadrant  is  1,  what  is  the  diameter  of  the 
circle  ? 


Ex.  52.  Prove  that  the  area  of  the  square  inscribed  in 
a  sector  whose  central  angle  is  a  right  angle,  is  equal  to  one 
half  the  square  on  the  radius. 

Ex.  53.  If  a  circle  is  circumscribed  about  a  right 
triangle,  and  on  each  of  the  legs  of  the  triangle  as 
diameters  semicircles  are  drawn,  exterior  to  the 
triangle,  the  sum  of  the  areas  of  the  crescents  thus 
formed  equals  the  area  of  the  triangle. 

Prove  area  AECG  +  area  BFCH=  area  A  ABC. 

Suggestion.  —From  the  sum  of  A  ABC  and  the  semicircles  on  ^Cand  BC, 
subtract  the  semicircle  on  AB.  Express  each  area  in  terms  of  sides  a,  b,  and 
c  of  the  triangle. 

Ex.  54.  Construct  three  equal  circles  having  the  vertices  of  an  equi- 
lateral triangle  as  their  centers  and  for  their  radii  one  half  the  side  of  the 
triangle.  Compute  the  area  of  that  part  of  the  interior  of  the  triangle 
which  is  exterior  to  each  of  the  circles,  if  the  length  of  the  side  of  the 
triangle  is  s. 

Ex.  55.  Upon  a  segment  AC  draw  a  semicircle.  Upon  AC  locate  a 
point  B,  not  the  center  of  AC.  Upon  AB  and  BC  a,s  diameters  draw 
semicircles  within  the  one  drawn  upon  ^C  as  diameter.  Prove  that  the 
area  of  the  surface  lying  within  the  largest  semicircle  and  exterior  to  the 
smaller  ones  equals  the  area  of  the  circle  drawn  upon  BD  as  diameter, 
where  BD  is  the  perpendicular  to  AC  at  B  meeting  the  largest  semicircle 
at  D.     (Due  to  Archimedes.) 

Ex.  56.  With  the  vertices  of  an  equilateral  triangle  as  centers  and 
the  side  of  the  triangle  as  radius,  three  equal  circles  are  drawn.  Deter- 
mine the  area  of  that  figure  which  is  common  to  the  three  circles. 


304 


SUPPLEMENTARY  EXERCISES 


Ex.  57.     Express  in  terms  of  the  radius  B  the  area  of  the  segment  of 
a  circle  whose  chord  is  a  side  of  the  inscribed  square. 

Ex.  58.     Repeat  Ex.  67  if  the  chord  is  the  side  of  the  equilateral 
inscribed  triangle. 

Ex.  59.     The  arch  ABC  is  a  lancet  arch.     It  consists  of  two  arcs  with 
equal  radii,  drawn  from  centers  Ci  and  d  out- 
side the  span  BG.     Within  the  arch  are  two 
other  lancet  arches. 

Let  JSO  =  2  a  ;  let  CiCa  =  s  ;  let  J5Z)  =  BC. 

(a)  Determine  the  height  h  of  the  arch  BAC. 

(b)  What  is  the  length  of  the  radius  of  the  arc 
XD? 

(c)  What  is  the  height  of  the  arch  BXD  ? 

(d)  What  is  the  radius  of  the  circle  indicated  as  tangent  to  the  arches  ? 

(e)  What  is  the  area  and  circumference  of  the  circle  ? 


Ex.  60.  In  a  given  circle,  inscribe  three  equal 
circles,  tangent  to  each  other  and  to  the  given 
circle. 


Ex.  61.  In  a  given  equilateral  triangle,  in- 
scribe three  equal  circles,  tangent  to  each  other 
and  each  tangent  to  one  and  only  one  side  of  the 
triangle. 


Ex.  62.   The  figure  below  at  the  left  is  a  quatrefoil. 


(a)  Construct  such  a  figure  based  upon  a  square  whose  side  is  2  in. 
(6)  What  is  the  length  of  the  curved  line  if  AB  =  s  inches  ? 


SUPPLEMENTARY   EXERCISES 


305 


(c)  What  is  the  area  within  the  curved  line  if  AB  =  s  inches  ? 

(d)  Notice  that  the  quatrefoil  is  used  in  the  adjoining  design. 

Ex.  63.   Construct  a  figure  like  Fig.  1,  below,  upon  a  square  of  side 
2  in. 


Fig.  1 


Fig.  2 


(a)  What  is  the  length  of  the  curved  line  when  the  side  of  the  square 
is  s  inches  ? 

(b)  What  is  the  total  area  within  the  curved  line  when  the  side  of  the 
square  is  s  inches  ? 

(c)  Notice  that  the  curved  line  of  Fig.  1  is  the  fundamental  unit  of  the 
adjoining  window  design. 


SOLID   GEOMETRY 


BOOK  VI 

LINES  AND  PLANES  — POL YEDKAL 
ANGLES 

442.  Surfaces.  No  satisfactory  elementary  definition  of 
surface  in  general  can  be  given.  The  surface  of  a  physical 
object  is  that  part  of  the  object  which  can,  in  general,  be 
touched;  it  separates  the  portion  of  space  occupied  by  the 
object  from  surrounding  space. 

The  surface  of  a  small  pond  on  a  calm  day  is  approximately  a,  plane 
surface. 

The  surface  of  a  croquet  ball  or  of  a  billiard  ball  is  a  spherical  surface. 
The  surface  of  a  "  round  "  marble  column  is  a  cylindrical  surface. 

Ex.  1.  If  two  points  on  the  surface  of  a  ball  were  joined  by  a  straight 
line,  where  would  the  line  lie  ? 

Ex.  2.  Are  there  any  two  points  on  the  surface  of  a  ball  such  that  the 
straight  line  through  them  lies  upon  the  surface  of  the  ball  ? 

Ex.  3.  Are  there  two  points  on  the  surface  of  a  cylindrical  column 
such  that  the  straight  line  joining  them  lies  on  the  surface  of  the  column  ? 

Ex.  4.  Does  the  straight  line  joining  every  pair  of  points  on  the  sur- 
face of  a  cylindrical  column  lie  upon  the  surface  of  the  column  ? 

443.  A  Plane  is  a  surface  such  that  the  straight  line  join- 
ing any  two  points  of  it  lies  wholly  in  it. 


A  plane  is  represented  to  the  eye  by  a  quadri-  D 

lateral  like  the  figure  adjoining.  / 

The  plane  may  be  referred  to  as  plane  ABCD,  /, 

as  plane  AC,  or  as  plane  M.  A^ — 

307 


308  SOLID   GEOMETRY  —  BOOK  VI 

Pupils  will  find  it  convenient  at  times  to  represent  a  plane  by  a  thin 
card. 

Ex.  5.  In  plane  geometry,  it  was  agreed  that  a  straight  line  is  in- 
definite in  extent.  Do  you  think  we  should  agree  that  a  plane  is  indefinite 
in  extent  ?     Why  ? 

Ex.  6.     Let  MN  represent  a  thin  card  of  which  AB  is  an  edge.     Let 
C  represent  the  "  point "  of  a  pencil  lying  above 
MN.     Let  MN  be  turned  about  AB  as  an  axis  in 


the  direction  indicated  by  the  arrow\  ^"^^  q 

A       

Will  the  card  eventually  come  in  contact  with  / 

point  C?  / 

If  AB  and  C  are  kept  stationary,  will  the  card    B'  —^ 


come  in  contact  with  C  in  more  than  one  position  of  the  card  ? 

444.  A  plane  is  determined  by  a  combination  of  points  and 
lines  if  it  is  the  only  plane  which  contains  those  points  and 
lines. 

Points  and  lines  lying  in  the  same  plane  are  said  to  be 
Co-planar. 

445.  Postulate.     A  plane  can  he  extended  indefinitely. 

446.  Axiom.  A  plane  is  determined  by  three  non-collinear* 
points. 

Ex.  7.  Do  two  straight  lines  drawn  at  random  necessarily  lie  in  a 
plane  ?     Illustrate  by  holding  two  pencils. 

Ex.  8.  Do  four  points  usually  lie  in  a  plane  ?  Select  four  in  the  school- 
room that  do  and  four  that  do  not. 

Ex.  9.  Are  two  straight  lines  in  space  which  do  not  meet  no  matter 
how  far  they  are  extended  parallel  ? 

Ex.  10.  Why  is  a  tripod  used  as  mounting  for  a  camera  or  a  sur- 
veyor's instrument  ? 

Ex.  11.  Why  does  a  stool  with  three  legs  stand  firmly  whereas  one 
with  four  legs  cannot  always  be  made  to  stand  firmly  ? 

Ex.  12.  Prove  that  a  plane  and  a  straight  line  not  lying  in  the  plane 
can  have  only  one  common  point. 

*  Nou-collinear  points  are  points  which  do  not  all  lie  in  one  straight  line. 


LINES  AND  PLANES  309 

Proposition  I.     Theorem 

447.  A  plane  is  determined  by 

I.  A  straight  line  and  a  point  outside  the  line, 

II.  Two  intersecting  straight  lilies. 

III.  Two  parallel  straight  lines. 


1.  Hypothesis.     Point  C  lies  outside  st.  line  AB. 
Conclusion.     C  and  AB  determine  a  plane. 

Proof.     1.   Points   A,  B,  and    C  lie   in   one  and  only  one 

plane.  §  446 

2.  AB  lies  in  that  plane.  §  443 

3.  .*.  AB  and  C  determine  a  plane.  §  444 


II.     Hypothesis.     AB  and  BO  are  intersecting  st.  lines. 

Conclusion.     AB  and  BC  determine  a  plane. 

Proof.     1.   AB  and  point  C  determine  a  plane.  §  447,  I 

2.  BC  lies  in  that  plane.  §  443 

3.  .*.  AB  and  BC  determine  a  plane. 


III.     Hypothesis.     AB  and  CD  are  parallel  straight  lines. 

Conclusion.     AB  and  CD  determine  a  plane. 

Proof.     1.  AB  and  CD  lie  in  a  plane.  §  89 

2.  AB  and  CD  cannot  lie  in  more  than  one  plane,  for,  if  they 
did,  points  A,  B,  and  C  would  lie  in  more  than  one  plane,  which 
is  impossible.  Why  ? 

3.  .-.  AB  and  CD  determine  a  plane. 


310 


SOLID   GEOMETRY  —  BOOK  VI 


448.  The  Intersection  of  two  surfaces  or  of  a  surface  and  a 
line  consists  of  all  points  common  to  the  surfaces,  or  to  the 
surface  and.  the  line. 

449.  If  a  straight  line  intersects  a  plane,  the  point  of  inter- 
section of  the  line  and  the  plane  is  called  the  Foot  of  the  line. 

450.  Axiom.  If  two  planes  intersect,  they  have  at  least  two 
common  points. 


Proposition   II.     Theorem 
451.    The  intersection  of  tioo  planes  is  a  straight  line. 


Hypothesis.  A  and  B  are  two  points  common  to  planes  MN 
and  PQ. 

Conclusion.  The  intersection  of  MN  and  PQ  is  a  straight 
line. 

Proof.     1.  Draw  straight  line  AB. 

2.  AB  lies  in  plane  MN  and  also  in  plane  PQ.       Why  ? 

3.  No  point  outside  AB  can  be  in  both  MN  and  PQ,  for,  if 
there  were,  MN  and  PQ  would  coincide.  §  447,  I 

4.  Hence  the  complete  intersection  of  MN  and  PQ  is  the 
straight  line  AB.  '  §  448 

452.   A  line  is  perpendicular  to  a  plane  if  it  is  perpendicular 

to  every  line  in  the  plane  passing  through  its  foot. 
The  plane  is  also  perpendicular  to  the  line. 


LINES  AND  PLANES 


311 


Proposition   III.     Theorem 

453.  If  a  line  is  per2)endicular  to  each  of  two  inter- 
secting lines  at  their  intersection,  it  is  perpendicular  to 
their  plaiie. 


Hypothesis.     AD  and  AC  intersect  at  A^  determining  plane 
RS. 
BALAD^xi^BAl^AC. 
Conclusion.  BA  ±  plane  RS. 

Proof.     1.   Let  AE  be  any  other  straight  line  in  R8  through  A. 
Let  DQ  be  a  straight  line  in  R8  intersecting  AD^  AE,  and 
AC,  at  Z),  E,  and  (7,  respectively. 

2.  Extend  Bx\  to  B\  making  B'A  =  BA. 
Draw  BD,  BE,  BC,  B'D,  B'E,  and  B'C. 

3.  AD  and  AC  are  ±  bisectors  of  BB'.  Why  ? 

4.  .-.  ^Z)  =  i^'i>  and  BC  =  B'C.  Prove  it. 
6.                            .-.  A  iJZ>C  ^  A  5'Z)a  Prove  it. 

6.  Revolve  A  B'DC  on  DC  as  axis  until  ^'  falls  on  B. 

7.  Then  B'E  =  BE.  Why  ? 

8.  .'.A  and  JE7  are  both  equidistant  from  B  and  B'. 

9.  .-.  ^^  ±  BB',  or  5B'  ±  ^^.  §  77 

10.  But  ^£7  is  any  st.  line  in  iJ-S'  through  A. 

11.  .*.  ^B  ±  every  st.  line  in  RS  through  A,  and  hence 


AB±  plane  RS. 


§452 


Ex.  13.    If  a  line  is  perpendicular  to  a  line  of  a  plane,  is  it  perpendic- 
ular to  the  plane  ? 


312 


SOLID   GEOMETRY  —  BOOK  VI 


454.  Cor.  1.     Through   a  point   of  a 

line  a  plane  can  he  drawn  perpendicular 

to  the  line. 

Suggestion.  —  Draw  BC  and  BD,  any  two  per- 
pendiculars to  AB. 

Note.  —  Through  a  point  of  a  line  only  one  plane  can  he  drawn  per- 
pendicular to  the  line.  If  ME  and  BE  were  both  ± 
to  AB  at  J5,  a  plane  AD  through  AB  would  intersect 
ME  and  BE  in  two  lines  BF  and  jBZ>,  each  ±  to  ABT 
at  B.  But  this  is  impossible,  for,  in  a  plane  {AB)., 
only  one  line  can  be  drawn  perpendicular  to  a  given 
line  at  a  point  of  the  line. 

455.  Cor.  2.     Through  a  point  outside  a  line,      ^ 
a  plane  can  he  drawn  perpendicular  to  the  line.    (See  Fig.  §  454.) 

Suggestion.  — Draw  CB  1  AB  from  C;  then  draw  BD  1  AB  at  D. 

Note.  —  Through  a  point  outside  a  line,  only  one 
plane  can  he  drawn  perpendicular  to  the  line.  If 
planes  BT  and  8T  through  C  were  both  ±  to  AB., 
the  plane  ABC  determined  by  AB  and  C,  would 
intersect  BT  and  ST  in  lines  XC  and  YC,  each  ± 
to  AB.  But  this  is  impossible,  for,  in  a  plane.,  only 
one  line  can  be  drawn  perpendicular  to  a  given  line 
from  a  point  outside  the  line. 

456.  Cor.  3.     At  a  point  in  a  plane,  a  straight  line  can  he 
drawn  perpendicular  to  the  plane. 

Construction.     1.   Draw  CD  any  line  in 
iifiNT  through  O. 

2.  Draw  plane  BB  ±  to  CD  at  O,  meet- 
ing MN  in  line  AB. 

3.  Draw  EO  in  plane  BB,  ±  to  AB  at  O. 

Statement.  EO  ±  plane  MN  at  0. 

Note.  —  At  a  point  of  a  plane,  only  one  straight  line 
can  be  drawn  perpendicular  to  the  plane.  If  EO  and 
E'O  were  both  ±  to  plane  MN  at  O,  they  would  de- 
termine a  plane  BB  which  would  intersect  MN  in 
line  AB.  EO  and  E'O  would  both  be  ±  to  AB  at  0, 
and  that  is  impossible.     Why  ? 


I 


LINES  AND  PLANES 


313 


457.  Cor.  4.  Any  point  in  the  plane 
ichich  is  perpendicular  to  a  segment  at  its 
mid-point  is  equidistant  from  the  ends  of 
the  segment. 

Ex.  14.  Each  of  three  concurrent  lines  is 
perpendicular  to  each  of  the  other  two.  Prove 
that  each  is  perpendicular  to  the  plane  of  the  other  two. 

Ex.  15.  If  two  oblique  lines,  drawn  to  a  plane  from  a  point  in  a 
perpendicular  to  the  plane,  cut  oflE  equal  distances  from  the  foot  of  the 
perpendicular,  they  are  equal. 

Ex.  16.    State  and  prove  the  converse  of  Ex.  15. 

Ex.  17.  If  two  oblique  lines,  drawn  to  a  plane 
from  a  point  in  a  perpendicular  to  the  plane,  cut  off 
unequal  distances  from  the  foot  of  the  perpendicular, 
the  more  remote  is  the  greater. 

If  BE  >  BD,  prove  AE  >  AD. 
Suggestion.  —  Take  BF  =  BD,  and  draw  AF.  Recall  §  165 

Ex.  18.     State  and  prove  the  converse  of  Ex.  17. 
Suggestion.  —Give  an  indirect  proof,  basing  it  upon  Ex.  15  and  17. 

Ex.  19.  If  a  circle  be  drawn  in  a  plane  and  at  its  center  a  perpen- 
dicular to  the  plane  be  erected,  any  point  in  this  perpendicular  is  equi- 
distant from  the  points  of  the  circle. 

Ex.  20.  A  line  segment  of  fixed  length,  having  one  extremity  at  a 
fixed  point  lying  outside  a  plane,  has  its  other  extremity  in  the  plane. 
What  is  the  locus  of  the  extremity  which  lies  in  the  plane  ? 

Ex.  21.  IIow  many  different  planes  can  be  passed  through  one  straight 
line '? 

Ex.  22.    How  many  different  planes  are  determined  by  : 
(a)  Three  concurrent  lines  which  do  not  all  lie  in  one  plane  ? 
(6)  Three  parallel  lines  which  do  not  all  lie  in  one  plane  ? 
(c)  Two  intersecting  lines  and  a  point  which  does  not  lie  in  their  plane? 
{d)  Four  points,  no  three  of  which  are  collinear  and  which  do  not  all 
lie  in  one  plane  ? 

Ex.  23.  Prove  that  two  parallels  and  any  transversal  of  them  are 
co-planar. 

Ex.  24.  How  many  lines  of  intersection  are  determined,  in  general, 
by  three  planes  ? 


314 


SOLID   GEOMETRY  —  BOOK  VI 


Proposition   IV.     Theorem 

458.  All  the  perpendiculars  to  a  straight  line  at  a 
point  of  the  line  lie  in  a  plane  perpendicular  to  the  line 
at  the  point. 


Hypothesis.  AC,  AD,  and  AE  are  any  three  Js  to  AB  at  A. 
Conclusion.  AC,  AD,  and  AE  lie  in  a  plane  A.  to  AB  at  A. 
Proof.     1.     Let  AC  and  AD  determine  plane  MN. 

2.  .-.  xiB  A.  plane  MN.  Why  ? 

3.  Let  AB  and  AE  determine  plane  ABE,  intersecting  MN 
in  AE\ 

4.  .-.  AB  ±  AE,  since  AE^  is  in  MN.  §  453 

5.  .-.  AE  and  AE,  both  in  plane  BE,  must  coincide. 

[/)i  a  plane,  only  one  _L  can  be  drawn  to  a  line  at  a  point  in  the  line.] 

§81 

6.  .-.  AE  must  lie  in  MN.  Why  ? 

7.  Hence  all  Js  to  AB  at  J.  must  lie  in  MN.         Why  ? 


3f. 


459.  Cor.  1.  Any  point  equidistant  from 
the  eyids  of  a  seginent  lies  in  the  plane  per- 
pendicular to  the  segment  at  its  midpoint. 

460.  Cor.  2.  TJie  locus  of  points  in  space 
equidistant  from  the  ends  of  a  segment  is  the 
plane  perpendicular  to  the  segment  at  its  mid-point. 


D 


y-. 


Suggestion.— Reyiew,  if  necessary,  §  229  of  the  Plane  Geometry  and  apply 
§§  457  and  459. 


LINES  AND  PLANES 


315 


Proposition   V.     Theorem 

461.  If  through  the  foot  of  a  perpendicular  to  a 
plane  a  line  he  drawn  at  right  angles  to  any  line  in 
the  plane,  the  line  drawn  from  its  intersection  with  this 
line  to  any  point  in  the  perjyendicidar  will  he  perpen- 
dicular to  the  line  in  the  plane. 


Hypothesis.     AB  ±  plane  MN\    CD  is  any  line  in  MN\ 
AE  J_  CD  ;  BE  is  drawn  from  any  point  B  of  AB  to  E. 
Conclusion.  BE  A.  CD. 

Suggestions.  — 1.    Take  CE  =  DE,  and  draw  BD,  BC,  AD,  and  AC. 
2.    Compare  ^C  and  ^Z).  3.    Compare  5Z)  and  56'. 

462.  Cor.  1.  From  a  point  outside  a 
plane,  a  straight  line  can^be  drawn  per- 
pendicular  to  the  plane. 

Construction.  1.  Draw  DE,  any  st.  line  in 
MX. 

2.  Draw  AF  ±  to  DE  at  F,  and  BF,  in 
Jf  iV,   ±  to  DE  at  F. 

3.  Draw  AB  ±  to  BF. 
Statement.                   AB  ±  plane  MN  from  A. 
Proof.     1.  Draw  BE. 

2.  EF  ±  the  plane  determined  by  AF  and  BF.  Why  ? 

3.  .'.BE±AB.  §461 
[Since  BF,  through  the  foot  of  EF,  is  ±  to  AB  in  plane  ABF.^ 

4.  .•.AB±3IN. 

[See  step  3  of  the  Construction  and  of  the  Proof.] 

Note. —  From  a  point  outside  a  plane  only  one  straight  line  can  he 
drawn  perpendicular  to  the  plane.  If  AC  and  AB  were  both  ±  to  plane 
J/iVfrom  A,  A  ABC  would  have  two  right  angles  in  it. 


316  SOLID   GEOMETRY  — BOOK  VI 

463.  Cor.  2.     The  perpendicular  is  the  shortest  segmerit  that 
can  be  drawn  from  a  point  to  a  plane. 

464.  The  Distance  from  a  point  to  a  plane  is  the  length  of 
the  perpendicular  from  the  point  to  the  plane. 

PARALLEL  LINES  AND  PLANES 

465.  A  straight  line  is  parallel  to  a  plane  if  it  does  not  meet 
the  plane  however  far  they  are  extended. 

Two  planes  are  parallel  if  they  do  not  meet  however  far 
they  are  extended. 

Proposition  VI.     Theorem 

466.  If  a  line  outside  a  plane  is  parallel  to  a  line  of 
^  the  plane,  it  is  parallel  to  the  plane. 


Hypothesis.  AB  II  CD. 

Plane  MN  contains  CD  but  not  AB. 
Conclusion.  AB  II  plane  MN. 

Proof.     1.   AB  and  CD  lie  in  a  plane  AD.  §  89 

2.  This  plane  intersects  MN  in  line  CD.  Why  ? 

3.  If  AB  were  to  intersect  MN,  the  point  of  intersection 
would  be  in  plane  MN  and  also  in  plane  AD,  and  therefore 
in  CD.      ■ 

4.  Hence  AB  would  intersect  CD. 

5.  But,  AB  cannot  meet  CD.  Why  ? 

6.  .*.  AB  cannot  meet  JOT"  and  hence  AB  II  MW. 


PARALLEL  LINES  AND  PLANES 


317 


467.     Cor.   1.    TJirough  a  given  straight  line,  a  plane  can  be 
drawn  parallel  to  any  other  straight 
line. 

Prove  a  plane  can  be  drawn  through 
CD  II  AB. 

Suggestion.  —  Bra.w  CE  \\  AB. 

Note.  —  Discuss    the    solution    when 
AB  II  CD  and  when  AB  is  not  II  CD. 


A 

B 

M 

^C 

T1            / 

/ 

/     '^ 

^^^^  / 

N 

468.   Cor.   2.    Through  a  given  point,  a  plane  can  he  drawn 
parallel  to  each  of  two  given  straight 
lines  in  space. 

Prove  a  plane  can  be  drawn  through 
point  A,  parallel  to  straight  lines  BC 
and  DE. 


Suggestion.  —  DTa\r  AG  \\  DE  and  AF 
BC. 


B- 

D\ 

c 

M 

^^E 

A                     ^  / 

y 

\ 

\(?  / 

N 

Note.  —  Discuss  the  solution  when  BC  II  2)^  and  when  BC  is  not  II  DE. 

Ex.  25.     Can  more  than  one  perpendicular  be  drawn  to  a  line  at  a 
point  of  the  line  in  a  plane  ?   in  space  ? 

Ex.  26.   Through  a  given  point  outside  a  line,  a  plane  can  be  drawn 
parallel  to  the  line.     How  many  such  planes  can  be  drawn  ? 

Ex,  27.  I*rove  that  a  straight  line  and  a  plane, 
both  perpendicular  to  the  same  straight  line,  are 
parallel. 

AB±AC;  plane  MN'±  A  C. 


Hyp. 
Con. 


AB  II  plane  MN. 


7 


Suggestion.  —  Let  the  plane  determined  by  AB  and 
AC  intersect  MN  in  line  DE. 

Ex.  28.  Through  a  given  point  outside  a  plane,  a  line  can  be  drawn 
parallel  to  the  plane.  Can  more  than  one  line  be  drawn  parallel  to  the 
plane  ? 

Ex.  29.  Through  a  given  point,  a  line  can  be  drawn  parallel  to  each 
of  two  intersecting  planes. 


318  SOLID   GEOMETRY  —  BOOK  VI 

Pkoposition  VII.     Theorem 

469.  If  a  straight  line  is  parallel  to  a  plane,  the 
intersection  of  the  plane  tvith  any  plane  draivn  through 
the  line  is  parallel  to  the  line. 


Hypothesis.  AB  li  plane  MJSf. 

Plane  BCj  through  AB,  intersects  JfiV  in  CD. 

Conclusion.  AB  II  CD. 

Proof.     1.   AB  and  CD  lie  in  the  same  plane  BC. 

2.  AB  and  CD  cannot  intersect,  for  if  they  did,  AB  would 
intersect  plane  MN,  which  is  impossible. 

3.  .-.  AB  II  CD.  §  89 

470.   Cor.   If  a  Uyie  and  a  plane  are  parallel,  a  parallel  to  the 
line  through  any  point  of  the  plane  lies  in  the  plane. 

Hyp.  AB  II  plane  MN. 

C  is  any  pt.  in  MN.  -^I 1^ 

CI)  II  AB.  /         / 

Con.  CD  lies  in  MN.  i + — ^/ 7^" 

Proof.     1.   The  plane  determined  by  AB  and         /    ^ ^    / 

G  intersects    MN  in   a    line    CE.,    through    C,      -^^ 
parallel  to  AB.  Why  ? 

2.  But  CD,  through  O,  II  AB.  Why  ? 

3.  .'.CD  and  CE  coincide.  Why  ? 

4.  .'.  CD  lies  in  plane  MN. 

Ex.  30.   Prove  that  three  non-concurrent  straight  lines,  each  of  which 
intersects  the  other  two,  lie  in  a  plane. 


.   PARALLEL  LINES  AND  PLANES       319 

Proposition  VIII.     Theorem 

471.   If  two  parallel  planes  are  cut  hy  a  third  plane, 
the  intersections  are  parallel. 


31 

n 

// 

^\ 

^ 

P    ^ 

\ 

V 

c 

Q 

Hypothesis.        Plane  MN  li  plane  PQ. 

Plane  AD  intersects  MN  in  AB  and  PQ  in  CD. 

Conclusion.  AB  II  CD. 

Suggestion.  —  Recall  §  89. 

Ex.  31.   Prove  that  parallel  segments  between  parallel  planes  are  equal. 

Ex.  32.  If  two  planes  are  parallel,  a  line  parallel  to  one  of  them 
through  any  point  of  the  other  lies  in  the  other.     (Fig.  of  Prop.  VIII.) 

Suggestion.  —  Given  parallel  planes  MN  and  P  Q,  and  AB  through  any 
point  A  of  MN  \\  PQ.  Prove  that^J?  lies  in  MN.  Through  AB  pass  a  plane 
intersecting  MN  in  a  line  AB'  and  PQ  in  line  CD.  Consider  the  relation  of 
lines  AB'  and  CD,  and  also  of  AB  and  CD. 

Ex.  33.  From  a  line  parallel  to  a  plane,  two  parallels  are  drawn  to 
the  plane  and  terminated  by  the  plane.    Prove  that  the  segments  are  equal. 

Review  Exercises 

Ex.  34.    When  is  a  line  perpendicular  to  a  plane  ? 

Ex.  35.  Where  do  all  lines  lie  which  are  perpendicular  to  a  given 
line  at  a  given  point  of  the  line  ? 

Ex.  36.  What  is  true  about  two  lines  in  the  same  plane  which  are 
perpendicular  to  the  same  line  ? 

Ex.  37.  Line  AB  is  perpendicular  to  plane  iHf.V  at  A.  A  line  is 
drawn  from  B  meeting  any  line  CB  of  plane  MN  at  E.  If  line  BE  is 
perpendicular  to  CD,  prove  AE  perpendicular  to  CD.     (Fig.  Prop.  V.) 


320 


SOLID    GEOMETRY  — BOOK  VI 


Peopositiois'   IX.     Theorem 

472.    Two  lines  perpendicular  to  the  same  plane  are 
parallel. 


A 

\ 

G 

M 

/  / 

B 

/ 

D     / 

Hypothesis.    AB^MN^\.B',  CDI.MN2XD. 
Conclusion.  AB  II  CD. 

Proof.     1.   Draw  AD  from  any  pt.  A  of  AB.     Draw  BD. 
In  plane  MN,  draw  FD  ±  BD. 

2.  CD1.FD.  Why? 

3.  AD  ±  FD.  §  461 

4.  .-.  AD,  BD,  and  CD  lie  in  a  plane  ABDG.  §  458 

5.  .'.  AB  and  CD  lie  in  the  same  plane,  ABDC. 

6.  But  AB  A.  BD  and  CD  ±  BD.  Why  ? 

7.  .*.  ^J5  II  CD.  §  97 


473.   Cor.  1.     If  one  of  two  parallel  lines  is  perpendicular  to  a 
plane,  the  other  is  also  perpendicular  to  the  plane. 

A        O 


Hyp. 


Con. 


AB  II  CD. 

AB  ±  plane  MJST. 

CD  ±  MN. 


Suggestions.  —  1.  Assume  CE  1  MN. 
2.  Prove  CE  \\AB. 

474.  Cor.  2.  If  each  of  two  straight 
lines  is  parallel  to  a  third  straight  line, 
they  are  parallel  to  each  other. 

Hyp.  ^J5II  CD;  EFW  CD. 

Con.  AB  II  EF. 

Suggestion.  —  Let  plane  MNhe  1  CD. 


I    \     /I    /^ 

/       BED/ 
M'- ' 


PARALLEL  LINES  AND  PLANES  321 

Ex.  38.  Through  a  given  line  which  is  parallel  to  a  given  plane,  a 
number  of  planes  are  passed  intersecting  the  given  plane.  Prove  that 
the  lines  of  intersection  with  the  given  plane  are  all  parallel. 

Ex.  39.  If  two  parallel  planes  intersect  two  other  parallel  planes,  the 
four  lines  of  intersection  are  parallel. 

Ex.  40.  Prove  that  a  line  parallel  to  a  plane  is  everywhere  equidis- 
tant from  it. 

Ex.  41,  If  two  points  are  equidistant  from  a  plane,  and  on  the  same 
side  of  it,  they  determine  a  line  parallel  to  the  plane. 

Ex.  42.  If  two  points  lie  on  opposite  sides  of  a  plane  and  equidistant 
from  it,  the  segment  joining  them  is  bisected  by  the  plane. 

Ex.  43.  If  one  of  two  parallel  lines  is  parallel  to  a  plane,  the  other  is 
also,  unless  it  lies  in  the  plane. 

Suggestion.  —  Through  the  line  which  is  1|  to  the  plane,  pass  a  plane  inter- 
secting the  given  plane.    Use  §  466. 

Ex.  44.  Prove  that  the  lines  joining  in  order  the  mid-points  of  the 
sides  of  a  quadrilateral  in  space  form  a  parallelogram. 


Proposition   X.     Theorem 

475.    Two  planes  perpendicular  to  the  same  straight 
line  are  parallel. 

My 


N 


Hypothesis.     Planes  MN  and  PQ  are  _L  to  AB, 
Conclusion.  MN  \\  PQ. 

Suggestion.  — Prove  it  by  the  indirect  method,  using  §  455. 

Ex.  45.     Are  lines  in  space  which  are  perpendicular  to  the  same  line 
necessarily  parallel  ? 


322 


SOLID   GEOMETRY  — BOOK  VI 


Propositiois'   XI.     Theorem 

476.    If  each  of  tivo  intersecting  lines  is  j^ctrallel  to  a 
plancj  their  plane  is  pai^allel  to  the  given  plane. 


Hypothesis.     AB  II  plane  PQ ;  AC  11  plane  PQ, 
AB  and  AO  determine  plane  MN. 
Conclusion.  MN  II  PQ. 

Proof.     1.  .  From  A  draw  AD  ±  PQ. 

2.  AC  and  AD  determine  a  plane  which  intersects  PQ  in  a 
line  DF  parallel  to  AC.  Why  ? 

3.  Similarly,  DE  II  AB. 

4.  AD  A.  DF  and  also  AD  ±  DE.  Why  ? 

5.  .-.  ^Z>  ±  AC  and  also  J.Z>  ±  AB.  Why  ? 
Complete  the  proof,  using  §  475. 


Proposition^   XII.     Theorem 

477.   A  straight  line  perpendicular  to   one  of  tioo 
parallel  planes  is  pferpendicular  to  the  other  also. 

Hypothesis.     Plane  MN  II  plane  PQ.     AD  A.  PQ. 

(Fig.  §  476.) 
Conclusion.  AD  A.  MN. 

Proof.     1.    Through  AD  pass  two  planes  intersecting  MN 
in  AB  and  ACy  and  PQ  in  DE  and  DF,  respectively. 

2.  .-.  AB  II  DE  and  AC  II  J9i<^.  Why  ? 

3.  ^D  J.  i)£;  and  also  AD  ±  Di^.  Why  ? 

•  Complete  the  proof. 


/ 

-/ 

JS 

p 

/ 

'  / 

y 

PARALLEL  LINES  AND  PLANES  323 

478.  The  distance  between  two  parallel  planes  is  the  length 
of  the  segment  perpendicular  to  them  and 
lying  between  them. 

479.  Cor.  1.     Two  parallel  planes  are 
everywhere  equally  distant. 

480.  Cor.  2.     Through  a  given  point,  a 
plane  can  be  drawn  parallel  to  a  given  plane. 

Note.  —  TTirough  a  given  point  only  one  plane  can  he  draion  parallel 
to  a  given  plane.  For,  if  two  planes  through  A  were  parallel  to  PQ, 
each  would  be  ±  to  AB  at  A,  and  that  is  impossible. 

Proposition  XIII.     Theorem 

481.  If  two  angles  not  in  the  same  plane  have  their 
sides  parallel  and  extending  in  the  same  direction,  they 
are  equal,  and  their  planes  are  parallel. 


M 

/  iH\ 

/ 

/    ^iT" 

1.         / 

.V 

A  \ 

\ 

/\^ 

-^\    / 

/  y-^ 

^c'X 

Q 

Hypothesis.   Z  BACis  in  plane  MN;  Z  B'A!C'  is  in  plane  PQ. 

AB  II  A'B^  and  extends  in  the  same  direction. 

AC  II  A'C  and  extends  in  the  same  direction. 
Conclusion.     Z  BAC  =  Z  B'A'C ;  MN  II  PQ. 
Proof.     1.        AB  II  plane  PQ  and  AC  I!  PQ.  §  466 

2.  .-.  MN  II  PQ. 

3.  Lay  off  AB  =  AB',  and  AC  =  A'C  ;    draw  BC,  B'C, 
AA',  BB',  and  GO. 

4.  ABB:A' i^d.  O, QXidi. '.  BBf  ^AA^iTidiBB' WAA'.    Why? 
6.  Similarly,  CC  =  AA  and  CC  II  ^^.' 

6.  .-.  BB'C'C  is  a  O.  Prove  it. 

Complete  the  proof. 


324 


SOLID   GEOMETRY  —  BOOK   VI 


Proposition  XIV.     Theorem 

482.  If  two  straight  lines  are  cut  by  three  or  more 
parallel  planes,  the  corresponding  segments  are  pro- 
portional. 


Hypothesis.     Planes  MN,  PQ,  and  RS  are  parallel. 
AC  intersects  the  planes  at  A,  B,  and  C  respectively. 
A^C  intersects  the  planes  at  A\  B',  and  C  respectively. 

Conclusion.  4^  =  4^. 

BC     B'C 

Proof.     1.  Draw  AC  cutting  PQ  at  D, 

2.  Plane  CAC  intersects  PQ  at  BD  and  RS  at  CC 
Plane  AC  A'  intersects  PQ  at  DB'  and  MN  Sit  AA'. 

3.  .-.  BD  II  CC  and  also  BB'  \\  AA',  Why  ? 
Complete  the  proof,  using  §  265. 

Ex.  46.   Discuss  Prop.  XIII :  (a)  if  BA  and  B'A'  extend  in  opposite 
directions  and  CA  and  C'A.'  in  the  same  direction  ; 
(&)  if  both  pairs  extend  in  opposite  directions  from 
their  vertices. 

Ex.  47.  If  each  of  two  intersecting  planes  be 
cut  by  two  parallel  planes,  not  parallel  to  their  in- 
tersection, their  intersections  with  the  parallel 
planes  include  equal  angles. 

Suggestion.  — FroYB  Z  ABC=  Z  DBF. 

Ex.  48.     If  two  planes  are  parallel  to  a  third  plane,  they  are  parallel 
to  each  other.    (§  477  and  §  475.) 


DIEDRAL  ANGLES 


325 


ZDZ7 


DIEDRAL  ANGLES 

483.  It  is  evident  that  a  straight  line  divides  a  plane  into 
two   parts,   each  indefinite   in   extent.     A  ^ 

part  of  the   plane,  like  BCD,  is  called   a 
Half-plane.    BG  is  called  the  Edge  of  the    ^, 

half-plane.  ^  ^ 

484.  A  Siedral  Angle  is  the  figure  formed  by  two  half-planes 
which  have  a  common  edge. 

The  common  edge  is  the  Edge  and  the  two 
half-planes  are  the  Faces  of  the  diedral  angle. 

Thus,  half-planes  BF  and  BD  form  the  diedral  angle 
whose  edge  is  BE,  and  whose  faces  are  FBE  and  DBE. 

The  diedral  angle  may  be  read : 

diedral  Z  BE,  or  diedral  Z  ABEC. 

485.  Two  diedral  angles  are  Adjacent  when 
they  have  the  same  edge  and  a  common  face  be- 
tween them  :  as  Z  ABEC  and  Z  CBED. 

Two  diedral  angles  are  Vertical  when  the 
faces  of  one  are  the  extensions  of  the  faces  of 
the  other. 

486.  A  Plane  Angle  of  a  diedral  angle  is  the  angle  formed 
by  two  straight  lines  one  in  each  plane,  drawn  perpendicula.r 
to  the  edge  at  the  same  point. 

Thus,  if  lines  AB  and  ^O  be  drawn  in  faces  DE  and 
DF  respectively,  of  diedral  angle  DG,  perpendicular 
to  DO  at  A,  Z  BAC  is  a  plane  angle  of  the  diedral 
angle  DG. 


487.  Cor.  1.   All  p/ane  angles  of  a  given  die- 
dral angle  are  equal. 

488.  Cor.  2.  A  plane  loerpendicular  to  the  edge  of  a  diedral 
angle  intersects  the  faces  of  the  angle  in  lines  which  form  the 
plane  angle  of  the  diedral  angle. 

489.  Two  diedral  angles  are  equal  if  they  can  be  made  to 
coincide. 


326 


SOLID    GEOMETRY  —  BOOK   VI 


Proposition  XY.     Theorem 

490.    Tioo  diedral  angles  are  equal   if  their  plane 
angles  are  equal. 


Hypothesis.  Z  ABC  and  Z  A'B'C  are  the  plane  A  of  die- 
dral A  BD  and  B'D'  respectively ;   Z  ABC  =  Z  A'B'C 

Conclusion.  A  BD  =  A  B'D'. 

Proof.  1.  Apply  A  B'D'  to  Z  BD  so  that  A'B'  will  coincide 
with  AB  and  5'(7'  will  coincide  with  BC. 

2.  5i)  ±  plane  ABC,  and  JB'D'  ±  plane  A'B'C.  Why? 

3.  .-.  B'D'  coincides  with  BD.  Note,  §  456 

4.  .-.  ^'D'  coincides  with  AD,  and  0'i>'  with  CD.     §  447,  ll 

5.  .-.  Z  J5'Z>'  =  ABD.  §  489 

491.  Cor.  1.  i/*  tico  diedral  angles  are  equal,  their  plane  angles 
are  equal. 

For  the  diedral  angles  can  be  made  to  coincide.  Then  a  plane,  per- 
pendicular to  the  common  edge,  will  intersect  in  each  its  plane  angle,  and 
evidently  these  plane  angles  coincide. 

492.  Cor.  2.  If  two  planes  intersect,  the  vertical  diedral  angles 
are  equal. 

Suggestion.  —  Compare  their  plane  angles. 

Ex.  49.  Prove  that  a  plane  can  be  drawn  bisecting  a  diedral  angle. 

493.  A  diedral  angle  is  right,  acute,  or  obtuse  according  as 
its  plane  angle  is  right,  acute,  or  obtuse.  Two  diedral  angles 
are  supplementary  or  complementary  according  as  their  plane 
angles  are  supplementary  or  complementary. 


DIEDRAL  ANGLES 


327 


Ex.  50.  If  one  plane  meets  another  plane,  the  adjacent  diedral  angles 
formed  are  supplementary. 

Ex.  51.  If  two  parallel  planes  are  cut  by  a  third  plane,  the  alternate- 
interior  diedral  angles  are  equal. 

Suggestion.  —  Prove  the  plane  A  of  the  alt. -int.  diedral  A  equal. 

p 

494.  Two  planes  are  perpendicular  if  the 

diedral   angles   formed    are    right   diedral 
angles. 

Proposition  XVI.     Theorem 

495.  If  a  straight  line  is  perpendicular  to  a  plane, 
every  plane  drawn  through  the  line  is  perpendicular  to 
the  plane. 


Hypothesis.  AB  _L  plane  MN. 

PQ  is  any  plane  through  AB. 

Conclusion.  PQ  ±  MN. 

Proof.     1.   Let  PQ  and  MN  intersect  in  line  QR. 

2. 

3. 

4. 

6. 


Draw  C'BC  in  MN±  QR  at  B. 

AB±QR.  Why? 

.-.  Z  ABC  is  the  plane  angle  of  the  diedral  Z  PQRN.    Def. 
But  Z  ABC  is  a  rt.  Z.  Why  ? 

Complete  the  proof,  recalling  §§  494  and  493. 


Ex.  52.     (a)  Prove  that  a  plane  can  be  drawn  through  a  point  per- 
pendicular to  a  given  plane. 

(b)  How  many  such  planes  can  be  drawn  ? 

Ex.  53.     Prove  that  a  plane  perpendicular  to  the  edge  of  a  diedral 
angle  is  perpendicular  to  the  faces  of  the  angle. 


328  SOLID   GEOMETRY  —  BOOK  VI 

Propositiois^  XVII.     Theorem 

496.  If  two  planes  are  perpendicular  to  each  other,  a 
straight  line  drawn  in  one  of  them  perpendicular  to 
their  intersection  is  perpendicular  to  the  other. 


M^ 


C'A 


Why? 
Why? 


Q  N 

Hypothesis.     Plane  PQ  ±  plane   MN.     PQ  intersects  MN 

in  QR.     Line  AB  in  PQ  is  ±  QR. 
Conclusion.  AB  ±  MN. 

Proof.     1.  Draw  OBC  in  plane  MN  1.  QR. 

2.  .'.  Z  ABC  is  the  plane  Z  of  diedral  Z  PQRN. 

3.  But  ZPQi^  2V^  is  art.  diedral  Z. 

Complete  the  proof. 

497.  Cor.  1.  If  two  planes  are  perpendicular  to  each  other,  a 
perpendicular  to  one  of  them  at  a7iy  point  of  their  intersection  lies 
in  the  other. 

Hyp.  Plane  PQ  ±  plane  MN,  intersecting  it  in  QB. 

AB,  drawn  from  any  pt.  B  of  QB,  is  ±  to  MN. 
Con.  AB  lies  in  PQ. 

Suggestions.  —  1.   A  line  in  PQ  1  Q7?  at  £  is  1  MN.  Why  ? 

2.   Prove  that  it  and  AB  coincide. 

498.  Cor.  2.  If  two  planes  are  perpendicular  to  each  other,  a 
perpendicular  to  one  of  them  from  any  poi7it  of  the  other  lies,  in 
the  other. 

Hyp.  Plane  PQ  ±  plane  MN,  intersecting  it  in  QB  ; 

AB,  drawn  from  any  pt.  A  of  PQ,  is  ±  to  MN. 
Con.  AB  must  lie  in  PQ. 

Suggestions.  — In  PQ  draw  a  1  to  QR  from  A.  Prove  that  AB  must  coin- 
cide with  this  perpendicular. 


DIEDRAL  ANGLES 


329 


Proposition  XVIII.     Theorem 

499.   A  plane  perpendicular  to  each  of  tivo  intersect- 
ing planes  is  perpendicular  to  their  intersection. 


Hypothesis.     Plane  MN  A.  planes  RS  and  PQ. 
RS  intersects  PQ  in  line  AB. 

Conclusion.  MN  ±  AB. 

Suggestion.  —  Assume  a  line  1  MN  from  A.    Where  will  this  line  lie  ? 

(§498.) 

Ex.  54.     Are  two  planes  which  are  perpendicular  to  the  same  plane 
necessarily  parallel  ? 

Ex.  55.     If  a  plane  is  perpendicular  to  a  line  of  a  plane,  it  is  perpen- 
dicular to  the  plane.  ^ 

Ex.  56.  If  a  straight  line  is  parallel  to  a  plane, 
any  plane  perpendicular  to  the  line  is  perpendicular 
to  the  plane. 

Hyp.  AB  II  plane  MN. 

Plane  PB  ±  AB  at  C. 

Con.  PB  ±  MN. 


Suggestions.  —  1.  Draw  line  CD  in  PR  1  QR. 

2.  Let  the  plane  determined  by  CB  and  CD  intersect  MNin  line  DE. 

3.  Prove  CD  1  MN. 

Ex.  57.  Prove  that  any  point  in  the 
plane  through  the  bisector  of  an  angle  and 
perpendicular  to  the  plane  of  the  angle  is 
equidistant  from  the  sides  of  the  angle.  j^- 

Suggestions.  —  I.  Draw  Pi?  1  OC,  RT 1  OA,  and  RS  L  OB.    Draw  PT 
and  PS. 

2.  Prove  PR  1  MN  (§  496),  PT  1  OA  (§  461),  PS  1  OB. 


330 


SOLID   GEOMETRY  —  BOOK  VI 


Pkoposition  XIX.     Theorem 

500.   Every  point  in  the  plane  hisecting  a  diedral 
angle  is  equidistant  from  the  faces  of  the  angle. 


Hypothesis.     Plane  BE  bisects  diedral  Z  ABDC. 
P  is  any  point  in  plane  BE. 
PM ±  plane  AD',  PN ±  plane  DC. 
Conclusion.  PM  =  PN. 

Proof.     1.   The  plane  determined  by  PM  and  PN  intersects 
planes  AD,  BE,  and  CD  in  lines  FM,  FP,  and  FN  respectively. 

2.  Plane  PMFJSf  ±  AD  and  also  ±  CD.  Why  ? 

3.  .-.  PMFN  ±  BD.  Why  ? 

4.  .-.  APFMsiiid  PEN  are  the  plane  A  of  diedral  AABDE 
and  CBDE.  Why  ? 

5.  .'.APFM=APFN.  Why? 

Complete  the  proof. 

501.   Cor.     Any  point  loithin  a  diedral  angle  and  equidistant 
from  its  faces  lies  in  the  plane  bisecting  the  diedral  angle. 

Suggestions.  —  1.  Let  BE  be  the  plane  determined  by  P  and  BD. 
2.  Prove  that  BE  bisects  Z  ABDC  by  proving  A  PFM  and  PEN  are  the 
plane  angles  of  the  diedral  angles  and  are  equal. 

Ex.  58.  If  perpendiculars  are  drawn  to  the  faces 
of  a  diedral  angle  from  any  point  within  the  angle, 
they  lie  in  a  plane  perpendicular  to  the  edge  of  the 
diedral  angle  and  form  an  angle  which  is  the  supple- 
ment of  the  plane  angle  of  the  diedral  angle. 

Suggestion.  — Wha,t  is  the  sum  of  the  angles  of  a 
plane  quadrilateral? 


DIEDRAL  ANGLES 


331 


Proposition  XX.     Theorem 

502.  Tlirough  a  given  straight  line  not  jjerpendicular 
to  a  given  2^lcine,  one  and  only  one  plane  can  he  drawn 
peypendicular  to  the  given  plane. 


AB  is  not  ±  to  plane  MN. 
A  plane  can  be  drawn  tlirough  AB  ±  MN,  and 


Hypothesis. 
Conclusion. 

only  one. 

Proof.     1.        Draw  AC  1.  MN,  from  point  A. 

2.  AC  and  AB  determine  a  plane  ±  MN.  Why  ? 

3.  If  a  second  plane  through  AB  were  J_  MN,  their  inter- 
section AB  would  also  be  X  MN.  Why  ? 

4.  But  AB  is  not  ±  MN. 

5.  .*.  Only  one  plane  can  be  drawn  through  AB  A.  MN. 

Note  1.  — If  AB  lies  in  MN,  the  theorem  is  still  true. 
Note  2. — If  AB  ±  MN,  an  infinity  of  planes  can  be  drawn  through 
AB±MN{^495). 

503.  The  projection  of  a  point  on  a  plane  is  the  foot  of  the 
perpendicular  drawn  from  the  point  to  the 
plane. 

The  projection  of  a  given  line  on  a  plane 
is  the  line  which  contains  the  projections 
of  all  the  points  of  the  given  line. 

Thus,  A'B'C  is  the  projection  of  ABC  on  MN. 

504.  Cor.     The  projection  of  a  straight  line  on  a  plane  is  a 
straight  line.     (Fig.  §  502.) 


Suggestions.  —  1.  Through  AB  pass  a  plane  AD  1  MN. 
2.  Prove  that  the  feet  of  the  J2  to  MN  from  AB  lie  in  CD. 


(§  498.) 


332  SOLID   GEOMETRY  —  BOOK  VI 

Proposition  XXI.     Theorem 

505.  The  acute  angle  between  a  straight  line  and  its 
projection  on  a  plane  is  the  least  angle  ivhich  it  makes 
ivith  any  line  drawn  in  the  plane  through  its  foot. 


Hypothesis.     BC  is  the  projection  of  AB  on  plane  MN. 

BD  is  any  other  line  in  MN  through  B. 
Conclnsion.  Z  ABC  <  A  ABB. 

Suggestions.  —  1.   Take -BZ)  =  £C.    2.   Compare  J. Z)  and  ^C. 
3.  Then  compare  A  ABC  and  ABD,  recalling  §  167. 

506.  The  angle  between  a  line  and  a  plane  is  the  acute  angle 
made  by  the  line  with  its  projection  on  the  plane.  This  angle 
is  called  the  Inclination  of  the  line  to  the  plane. 

Ex.  59.  If  two  equal  segments  are  drawn  to  a  plane  from  a 
point  outside  the  plane,  they  make  equal  angles  with  the  plane. 

Ex.  60.    If  two  parallels  meet  a  plane,  they  make 
equal  angles  with  it. 

Suggestion.  —  Giwen   AB   ||    CD;     A  A'  1  JfiV,  and 
CC  1  'm2^.    Prove  Z  ABA'  =  ACDC'. 

Ex.  61.  Prove  that  a  straight  line  and  its  projection  upon  a  plane  lie 
in  a  plane  which  is  perpendicular  to  the  given  plane. 

Ex.  62.  If  a  straight  line-segment  is  parallel  to  a  plane,  it  is  parallel 
to  its  projection  upon  the  plane,  and  is  equal  to  it. 

Ex.  63.  If  two  parallel  lines  are  oblique  to  a  plane,  their  projections 
upon  the  plane  are  parallel.     (§  481  and  §  471.) 

Ex.  64.  Prove  that  the  ratio  of  two  parallel  line-segments  is  the  same 
as  the  ratio  of  their  projections  upon  a  given  plane. 


u 

K 

/f 

// 

>/. 

Jc/ 

/  ^ 
/   ^ 

D 

/ 

-^N 

POLYEDRAL    ANGLES 


333 


Ex.  65.     Can  the  projection  upon  a  plane  of  a  curved  or  broken  line 
be  a  straight  line  ? 

Ex.  66.     If  the  projection  upon  a  plane  of  a  given  figure  is  a  straight 
line,  then  the  figure  lies  in  a  plane. 

Ex.  67.   Let  the  projection  upon  a  given  plane  M  of  a  segment  I  be  de- 
noted by  v.     What  is  tlie  relation  between  I  and  V  if  : 

(a)  Z  ±  itf  ?     {h)  IWM?     (c)  I  and  Jf  form  an  angle  of  45°  ? 

Note.  —  Supplementary  Exercises  1-17,  p.  454,  can  be  studied  now. 

POLYEDRAL  ANGLES 
507.   The  figures  below  represent  polyedral  angles. 
0  0  0 


B  C 

A  0-ABCDE 

Notice  that  each  is  formed  of  portions  of  three  or  more  inter- 
secting planes ;  these  planes  are  the  Faces  of  the  polyedral 
angle.  The  faces  intersect  in  one  common  point;  this  is  the 
Vertex  of  the  polyedral  angle.  Each  face  has  two  edges  which 
pass  through  the  vertex  of  the  angle ;  these  are  the  Edgesl  of 
the  polyedral  angle.  On  each  face,  the  two  edges  form  an 
angle,  whose  vertex  is  also  the  vertex  of  the  polyedral  angle  ; 
these  angles  are  the  Face  Angles  of  the  polyedral  angle.  Each 
pair  of  consecutive  faces  intersect  in  an  edge,  forming  a  diedral 
angle;  these  diedral  angles  are  the  Diedral  Angles  of  the 
polyedral  angle. 

The  edges  and  the  faces  are  unlimited  in  extent.  It  is  con- 
venient to  indicate  the  polygon  which  results  if  a  plane  is 
drawn,  not  through  the  vertex,  but  intersecting  all  the  faces ; 
this  polygon  aids  in  picturing  the  number  of  faces  of  the 
polyedral  angle. 


334  SOLID    GEOMETRY —  BOOK  VI 

508.  A  Triedral  Angle  is  a  polyedral  angle  having  three 
faces. 

A  Tetraedral  Angle  is  a  polyedral  angle  having  four  faces. 

Ex.  68.  If  a  polyedral  angle  has  4  faces,  how  many  vertices,  edges, 
face  angles,  and  diedral  angles  does  it  have  ? 

Ex.  69.  Name  the  edges,  face  angles,  and  the  diedral  angles  of  triedral 
angle  O-ABC.     (Fig.  §  507.) 

509.  Two  polyedral  angles  are  vertical  if  the  edges  of  one 
are  the  prolongations  of  the  edges  of  the  other. 

510.  Two  polyedral  angles  are  congruent  if  they  can  be  made 
to  coincide. 

Ex.  70.  (a)  Construct  two  triedral  angles  which  have  the  face  angles 
of  one  equal  respectively  to  the  face  angles  of  the  other  and  in  the  same 
order.     Determine  whether  they  can  be  made  to  coincide. 

(b)  Construct  a  third  triedral  angle  whose  face  angles  are  equal  to 
those  of  the  triedrals  of  part  (a),  but  arrange  them  in  order  opposite  to 
that  in  part  (a).  (See  Fig.  §511.)  Can  this  triedral  angle  be  made 
to  coincide  with  either  of  those  in  part  (a)? 

511.  Two  polyedral  angles  are  symmetrical  if  the  face  angles 
and  the  diedral  angles  of  one  are  equal  respectively  to  the  face 
angles  and  the  diedral  angles  of  the  other,  provided  these  parts 
occur  in  opposite  orders. 

Thus,  if  face  AAOB,  BOC,  and  COA  are  equal  respectively  to  face 
AA'O'B',  B'O'C,  and  CO' A',  and 

diedral  A  OA,   OB,  and   00  to  die-  0  O' 

dral  A  O'A',  O'B',  and  O'C,  triedral  /A  /v\ 

A  O-ABC  and  O'-A'B'C  are  sym-  //    \  /   \\ 

metrical,   since  the  parts   in  Z  0'-         /  \  /       \      \    / 

A'B'C  occur  in  opposite  order  to  the      A<^-h---^C       Ci^:^^--Ar--^A' 
equal  parts  of  Z  O-ABC;  that  is,  to  jb^  ^B^^ 

pass  from    OA  to   OB  to   OC,   one 

moves  from  left  to  right,  whereas,  to  pass  from  O'A'  to  O'B'  to  O'C,  one 
moves  from  right  to  left. 

It  is  evident  that,  in  general,  two  symmetrical  polyedrals 
cannot  be  placed  so  that  their  faces  will  coincide. 


POLYEDRAL  ANGLES 


335 


Proposition  XXII.     Theorem 
512.    Tioo  vertical  polyedral  angles  are  symmetricaL 


H3rpothesis.  0-ABC  and  O-A'B'C  are  vertical  triedral 
angles.      (Fig.  1.) 

Conclusion.     A  0-ABC  and  O-A'B'C  are  symmetrical. 

Proof.  1.  Face  A  AOB,  BOC,  etc.  equal  respectively  face 
AA'OB',  B'Oa,  etc.  Why  ? 

2.  Diedral  A  BOAC  and  B'OA'C  are  vertical ;  for  AOB  and 
A' OB'  are  parts  of  the  same  plane,  as  also  are  AOC  and  A' 00'. 

Similarly,  A  OB  and  OB'  are  vertical,  etc. 

3.  .•.  Diedral  A  OA,  OB,  etc.  equal  respectively  diedral 
angles  OA',  OB',  etc.  §  492 

4.  The  parts  of  0-ABC  occur  in  opposite  order  to  the  equal 
parts  of  Z  O-A'B'C. 

This  may  be  understood  by  moving  O-A'B'C  parallel  to  itself  to  the 
right,  and  then  revolving  it  about  an  axis  through  0  (as  shown  in  Fig.  2) 
until  face  OA'C  comes  into  the  same  plane  as  before.  OB'  is  then  in 
front  of  plane  C  OA'  instead  of  back  of  that  plane  as  in  Fig.  1.  Now,  in 
Fig.  1,  to  pass  from  ^0  to  OB  to  OC,  one  moves  from  left  to  right;  in 
Fig.  2  to  pass  from  OA'  to  OB'  to  OC ,  one  moves  from  right  to  left. 

5.  .-.  A  0-ABC  and  O-A'B'C  are  symmetrical.       §  511 

Note. — The  theorem  may  be  proved  for  any  two  vertical  polyedral 
angles  in  the  same  manner. 


336  SOLID   GEOMETRY  —  BOOK  VI 

Proposition  XXIII.     Theorem 

513.    The  sum  of  any  tioo  face  angles  of  a  triedral 
angle  is  greater  than  the  third. 

Note.  —  The  theorem  requires  proof  only  in  the  case  when  the  third 
face  angle  is  greater  than  either  of  the  others. 


Hypothesis.  In  triedral  Z  0-ABC 

Z.  AOC  >  Z  AOB,  and  also  Z  AOC  >  Z  BOC 
Conclusion.       Z  AOC  <ZAOB+Z  BOG. 
Proof.     1.    In  face  AOO,  draw  OD  =  OB,  making 

ZAOD  =  ZAOB. 
2.    Through  B  and  D  pass  any  plane  cutting  the  faces  of  the 
triedral  Z  in  AB,  BC,  and  CA,  respectively. 
3. 
4. 

5. 
6. 

7. 

8. 
9. 

Ex.  71.     Prove  that  any  face  angle  of  a  polyedral  angle  is  less  than 
the  sum  of  the  remaining  face  angles. 

Suggestion.  —  Divide  the  polyedral  Z  into  triedral  A  by  passing  planes 
through  any  lateral  edge,  and  apply  §  513. 


A  AOB  ^  A  AOD. 

Prove  it. 

.'.  AB  =  AD. 

Why? 

In  A  ABC 

AB  +  BO  >AD-{-  DO. 

Why? 

.'.BC>DC. 

§  158,  Ax.  18 

.'.  in  A  ^OC  and  A  COD 

ZBOC>ZCOD. 

§167 

AOB  +  Z  BOC  >  Z  AOD  +  Z  COD.        Why  ? 

.-.  Z  AOB  +  Z  BOC  >  Z  AOC. 

POLYEDRAL  ANGLES  337 

Proposition  XXIV.     Theorem 

514.    The  sum  of  the  face  angles  of  any  convex  poly e- 
dral  angle  is  less  than  four  right  angles. 


Hypothesis.     0-ABCDE  is  any  convex  polyedral  Z. 

Conclusion.     AAOB+  A  BOC  +  etc.  <  4  rt.  A. 

Proof.  1.  Pass  a  plane  cutting  the  faces  in  the  polygon 
ABCDE. 

Let  0'  be  any  point  within  ABCDE,  and  draw  OA,  O'B, 
O'G,  O'D,  and  O'E. 

2.  Then,  in  triedral  Z  A-EOB, 

Z  OAE  +  Z  OAB  >ZEAO'  -\-Z  O'AB.  §  513 

3.  Similarly  Z  OB  A  +  Z  OBC  >  Z  ABO'  +  Z  O'BC',  etc. 

4.  Adding  these  inequalities,  the  sum  of  the  base  A  of  the 
A  whose  common  vertex  is  0  is  greater  than  the  sum  of  the 
base  A  of  the  A  whose  common  vertex  is  0'.         §  158,  Ax.  19 

5.  The  sum  of  all  the  A  of  the  A  with  vertex  0  equals  the 
sum  of  all  the  A  of  the  A  with  vertex  0'.  ^Vhy  ? 

6.  .*.  the  sum  of  the  A  a^t  0  is  less  than  the  sum  of  the  A 
at  0'.  §  158,  Ax.  20 

7.  .-.  the  sum  of  the  z^  at  0  <  4  rt.  A.  Why  ? 

Note.  —  The  pupil's  understanding  of  this  theorem  will  be  increased  if 
a  pasteboard  model  of  the  figure  of  §  514  is  at  hand  when  this  theorem  is 
first  studied. 

Notice  that  the  inequality  in  step  2  does  not  mean  that  ZEAO 
>ZEAO'  and  Z  OAB  >  Z  O'AB ;  rather,  the  sum  of  Z  EAO  diud  Z  OAB 
is  greater  than  the  face  angle  EAB  of  triedral  Z  A-EOB. 


338 


SOLID   GEOMETRY  —  BOOK  VI 


Proposition  XXV.     Theorem 

515.  If  two  triedral  angles  have  the  face  angles  of 
one  equal  respectively  to  the  face  angles  of  the  other, 
their  homologous  diedral  angles  are  equal. 


C'C 


Fig.  1 


Fig.  2 


Fig.  3 


Hypothesis.     In  triedral  A  0-ABC  and  O'-A'B'C 
ZAOB  =  ZA'0'B'',  ZBOC=ZB'0'C'',  Z  00A  =  Z  C'O'A'. 

Conclusion.     Diedral  Z  OA  =  diedral  Z  OA';  etc. 

Proof.  1.  Lay  off  OA,  OB,  00,  O'A',  O'B',  and  O'C  all  of 
the  same  length,  and  draw  AB,  BO,  OA,  A'B',  B'C,  and  C'A'. 

2.  .-.  A  OAB  ^  A  O'A'B'  and  AB  =  A'B'.       Prove  it. 

3.  Similarly  BO  =  B'C  and  AO  =  A'O'. 

4.  AlsoAABC^AA'B'C'sindZBAO=ZB'A'C'.     Why? 

5.  On  OA  and  O'A',  take  AD  =  A'D' ;  draw  DE  in  face 
OAB±OA,  D'E'  in  A'0'B'±A'0',  DF  in  A0G1.A0,  D'F' 
in  A'0'0'  ±  A'O',  EF  in  face  ABC,  and  E'F'  in  face  A' B'C. 

6.  Then  A  ADE  ^  A  A' D'E'.  Prove  it. 
.-.  AE  =  A'E'  and  DE  =  D'E'. 

.  7.  Also  AF  =  A'F'  and  DF  =  D'F'.  Prove  it. 

8.  Then  A  AEF  ^  A  A' E'F'  and  EF  =  E'F'.    Prove  it. 

9.  Then  A  DEF  ^  A  D'E'F'.  Prove  it. 

10.  .-.  Z  EDF  =  Z  ^'i)'i^'.  Why  ? 

11.  .-.  diedral  Z  ^0  =  diedral  Z  ^'0'.  Why  ? 

Note. — The  above  proof  holds  for  Fig.  3  as  well  as  for  Fig.  2.  In 
Figs.  1  and  2,  the  equal  parts  occur  in  the  same  order,  and  in  Figs.  1  and 
3  in  the  reverse  order. 


SUPPLEMENTARY    TOPICS  339 

516.  Cor.  If  two  triedrcU  angles  have  the  face  angles  of  one 
equal  respectively  to  the  face  angles  of  the  other ^ 

1.  They  are  congruent  if  the  equal  parts  occur  in  the  same 
order. 

2.  They  are  symmetrical  if  the  equal  parts  occur  in  the  reverse 
order. 

SUPPLEMENTARY  TOPICS 

The  following  topics,  theorems,  and  exercises  of  Book  VI  are 
supplementary.  Each  group  is  independent  of  each  of  the  others. 
None  of  this  material  is  required  in  the  main  parts  of  subse- 
quent Books. 

Group  A.  —  Analogy  between  Triedral  Angles  and  Triangles. 

Group  B.  —  Loci  in  Space. 

Group  C.  —  Consists  of  two  supplementary  theorems  usually 
given  in  texts. 

Group  A.  Analogy  between  Triedral  Angles 
AND  Triangles 

517.  The  analogy  between  triangles  and  triedral  angles  is 
very  striking.  Many  propositions  of  plane  geometry  about  tri- 
angles may  be  changed  into  propositions  about  triedral  angles 
by  substituting  for  the  word  angle  of  the  former  diedral  angle, 
and  for  the  word  side  the  words /ace  angle. 

Ex.  72.  Two  triedral  angles  are  congruent  when  a  face  angle  and  the 
adjacent  diedral  angles  of  one  are  equal  respectively  to  a  face  angle  and 
the  adjacent  diedral  angles  of  the  other,  if  the  parts  are  arranged  in  the 
same  order. 

Suggestion.  —  Prove  by  superposition. 

Note.  —  If  the  parts  of  one  are  arranged  in  reverse  order  to  the  parts 
of  the  other,  the  triedral  angles  are  symmetrical. 

Ex.  73.  Two  triedral  angles  are  congruent  if  two  face  angles  and  the 
included  diedral  angle  of  one  are  equal  respectively  to  two  face  angles  and 
the  included  diedral  angle  of  the  other,  if  the  parts  are  arranged  in  the 
same  order. 


340  SOLID   GEOMETRY  —  BOOK  VI 

Ex.  74.  If  two  face  angles  of  a  triedral  angle  are  equal,  the  opposite 
diedral  angles  are  equal. 

Suggestion.  — Recsdl  the  proof  of  §  69. 

Ex.  75.  An  exterior  diedral  angle  of  a  triedral  angle  is  greater  than 
either  remote  interior  diedral  angle. 

Suggestion. —  Model  the  proof  after  that  in  §  87. 

Ex.  76.  —  If  two  triedral  angles  have  a  face  angle,  the  opposite  diedral 
angle  and  another  diedral  angle  of  one  equal  respectively  to  the  corre- 
sponding parts  of  the  other,  they  are  congruent  if  the  parts  are  in  the 
same  order. 

Suggestion.  —  Superpose  the  equal  face  angles,  so  that  the  equal  diedral 
angles  adjacent  to  the  faces  superposed  also  coincide.  Prove  that  the  faces 
opposite  these  diedral  angles  also  coincide  by  an  indirect  proof,  based  upon 
Exercise  75. 

Note.  —  If  the  parts  are  in  reverse  order,  the  figures  are  symmetrical. 

Ex.  77.     State  and  prove  the  converse  of  Ex.  74. 
Suggestion.  —  The  proof  is  based  upon  Ex.  76,  Note. 

Group  B.     Loci  in  Space 

518.  The  following  more  general  definition  of  locus  will  be 
employed  in  solid  geometry.  The  locus  of  points  satisfying  a 
given  condition  consists  of  all  points  which  satisfy  the  condi- 
tion, and.  of  no  other  points. 

The  points  which  constitute  a  locus  may  form  one  (or  more) 
lines,  or  one  or  more  surfaces. 

To  prove  that  a  particular  assumed  locus  is  actually  the  locus 
satisfying  a  given  condition,  or  conditions,  prove  either  (a)  and 
(b)  below  or  else  (a)  and  (c). 

(a)  Every  point  of  the  locus  satisfies  the  conditions. 

(h)  Every  point  not  of  the  locus  does  not  satisfy  the  conditions. 

(c)  Every  point  which  does  satisfy  the  conditions  lies  m  the 
locus. 

Ex.  78.'  As  a  consequence  of  §  457  and  §  459,  what  is  the  locus  of 
points  equidistant  from  the  extremities  of  a  line  ? 

Ex.  79.  As  a  consequence  of  §  500  and  §  501,  what  is  the  locus  of 
points  equidistant  from  two  intersecting  planes  ? 


SUPPLEMENTARY  TOPICS  341 

Ex.  80.  What  is  the  locus  of  points  in  space  at  a  given  distance  from 
a  given  plane  ? 

Ex.  81.  What  is  the  locus  of  points  in  space  equally  distant  from 
two  parallel  planes  ? 

Ex.  82.  What  is  the  locus  of  points  in  space  equidistant  from  the 
points  of  a  given  circle  ? 

Ex.  83.  What  is  the  locus  of  points  in  space  equidistant  from  the 
vertices  of  a  given  triangle  ? 

519.  Intersection  Loci.  —  When  two  conditions  are  imposed 
upon  a  point  in  space,  each  condition  determines  a  locus  for 
the  point,  and  the  desired  point  lies  in  the  intersection  of  the 
loci.  It  is  often  inadvisable  to  attempt  to  draw  the  two  loci, 
for  that  demands  considerable  skill  in  drawing.  The  following 
form  of  solution  brings  out  all  the  mathematical  value  of  such 
a  problem  quite  as  well,  if  not  better,  than  if  a  figure  were 
drawn. 

Illustrative  Problem.  —  What  is  the  locus  of  points  in 
space  at  a  given  distance  from  a  given  plane  and  equidistant  from 
two  given  points  ? 

Solution.  1.  The  locus  of  points  at  a  given  distance  from  a  given  plane 
consists  of  two  planes  parallel  to  the  given  plane  and  at  the  given  distance 
from  it.     Call  this  Locus  1. 

2.  The  locus  of  points  equidistant  from  two  given  points  is  the  plane 
perpendicular  to  and  bisecting  the  segment  between  the  two  points. 
Call  this  Locus  2. 

3.  The  desired  locus  is  the  intersection  of  Locus  1  and  Locus  2. 
Discussion.    1.    Generally  the  plane  of  Locus  2  will  intersect  the  planes 

of  Locus  1  in  two  straight  lines. 

2.  Locus  2  may  be  parallel  to  the  planes  of  Locus  1.  In  this  case, 
there  will  not  be  any  points  satifying  the  given  conditions. 

3.  Locus  2  may  coincide  with  one  of  the  planes  of  Locus  1.  In  this 
case,  the  plane  common  to  Loci  1  and  2  will  be  the  desired  locus. 

Ex.  84.  What  is  the  locus  of  points  in  a  given  plane  equidistant  from 
two  parallel  planes  ? 

Ex.  85.  What  is  the  locus  of  points  in  a  given  plane  at  a  given  dis- 
tance from  another  given  plane  ? 


342  SOLID   GEOMETRY  —  BOOK  VI 

Ex.  86.  What  is  the  locus  of  points  in  a  given  plane  equidistant 
from  two  given  points  not  in  the  plane  ? 

Ex.  87.  What  is  the  locus  of  points  in  a  given  plane  equidistant 
from  two  intersecting  planes  ? 

Ex.  88.  What  is  the  locus  of  points  equidistant  from  two  given  points 
and  also  equidistant  from  two  parallel  planes  ? 

Ex.  89.  What  is  the  locus  of  points  equidistant  from  two  given 
points  and  also  equidistant  from  two  intersecting  planes  ? 

Ex.  90.  What  is  the  locus  of  points  equidistant  from  two  parallel 
planes,  and  also  equidistant  from  two  intersecting  planes  ? 

Ex.  91.  What  is  the  locus  of  points  equidistant  from  two  intersect- 
ing planes,  and  also  at  a  given  distance  from  a  given  plane  ? 

Ex.  92.  Find  all  points  which  are  at  a  given  distance  from  a  given 
plane,  equidistant  from  two  other  parallel  planes,  and  equidistant  from 
two  given  points. 

Ex.  93.  Find  all  points  which  are  equidistant  from  two  given  inter- 
secting planes,  equidistant  from  two  parallel  planes,  and  equidistant  from 
two  given  points. 

Ex.  94.  Prove  that  the  three  planes  bisecting  the  diedral  angles  of 
a  triedral  angle  meet  in  a  common  straight  line.  o 

Snggestion.  —  Planes  OAD  and  OBE  intersect  in  /  i<t\ 

line  OG.    Prove  OG  is  in  plane  OCF.  /  //|1\\ 

Ex.  95.     Prove   that  the  three   planes   deter-        /      / liW  \ 
mined  by  the  edges  of  a  triedral  angle  and  the    ^^^^-:t-V-V::Y-^C 
bisectors  of  the  opposite  face  angles  intersect  in  a  f^^  m  /^ 

line.  ^B 

Suggestions.  —  1.  In  a  figure  like  that  for  Ex.  94,  assume  OB,  OF,  and 
OE  bisect  the  angles  BOO,  BOA,  and  AOC,  respectively. 

2.  Take  OA=OB=OC,  and  draw  AB,  BC,  and  AC 

3.  Prove  AD,  CF,  and  EB  are  concurrent. 

4.  Prove  planes  AOD,  BOE,  and  COF  meet  in  a  line. 

Ex.  96.  Prove  that  the  locus  of  points  equidistant  from  the  sides  of 
an  angle  is  the  plane  through  the  bisector  of  the  angle,  and  perpendicular 
to  the  plane  of  the  angle. 

Suggestion.  —  Recall  §  460  and  Ex.  57,  Book  VI. 

Ex.  97.  Prove  that  the  planes  through  the  bisectors  of  the  face 
angles  of  a  triedral  angle,  perpendicular  to  the  planes  of  the  faces,  meet 
in  a  line,  whose  points  are  equidistant  from  the  edges  of  the  triedral 
angle. 


SUPPLEMENTARY    THEOREMS 


343 


Group  C.     Supplementary  Theorems 

Proposition  XXVI.     Theorem 

520.  Tim  straight  lines  not  in  the  same  plane  have  one  com- 
mon perpendicular,  and  only  one  ;  and  this  segment  is  the  shortest 
segment  that  can  be  drawn  between  them. 


Hypothesis.     AB  and  CD  do  not  lie  in  the  same  plane. 

Conclusion.  One  and  only  one  common  ±  to  AB  and  CD 
can  be  drawn ;  also,  this  ±  is  the  shortest  segment  between 
AB  and  CD. 

Proof.     1.   Through  CD  draw  plane  MN  II  AB.  §  467 

2.  Through  AB,  draw  plane  AH  ±  plane  MN,  intersecting 
MN  in  line  OH.  §  502 

3.  .'.QHWAB.  Why? 

4.  .-.  GH  intersects  CD  at  a  point  G. 

[If  CD  II  GH,  CD  would  be  II  to  AB,  which  is  impossible.] 

5.  In  plane  AH,  draw  AG  ±  GH  at  G. 

6.  Then  AG  ±  AB  and  also  to  CD.  Prove  it. 

7.  Assume  KE  also  ±  to  both  AB  and  CD. 

8.  Draw  EF  II  AB,  and  KL  in  plane  AH  ±  GH. 

9.  ^ii^  is  in  MN. 

10.  EK±EF. 

11.  .'.EKA.MN. 

12.  But  this  is  impossible  for  /li  ±  307". 

13.  Hence  ^G^  is  the  only  common  X  to  AB  and  CD. 

14.  ^^  >  /O:. 

15.  .-.  EK  >  AG. 

16.  .-.  AG  is  the  shortest  segment  from  AB  to  CD. 


§470 
Why? 
Why? 

§496 
Why? 
Why? 
Why? 


344 


SOLID   GEOMETRY  —  BOOK   VI 


Proposition  XXVII.     Theorem 

521.    Two  diedral  angles  have  the  same  ratio  as  their  plane 
angles. 

Case  I.    When  the  plane  angles  are  commensurable. 


c 

Ill 

1    1    1 
1    1    1 

^ 

[^ 

Hypothesis.     A  ABC  and  A'B'C,  the  plane  A  of  diedral  A 
ABDC  and  A'B'D'C  respectively,  are  commensurable. 

Conclusion. = • 

Z  ABCD         Z ABO 

Proof.     1.    Let  Z  ABE,  a  common  measure  of  Z  ABC  and 
Z  A'B'C,  be   contained  4  times  in  Z  ABC  and   3   times  in 

Z  A'B'C. 

Z.  A'B'C     3 
4* 


2.    Then 


ZABC 

3.  Passing  planes  through  the  edges  BD  and  B'D',  and  the 
division  lines  of  ZABC  and  Z A'B'C,  respectively,  diedral 
Z  ABCD  is  divided  into  4  parts,  and  Z  A'B'D'C  into  three 
parts,  all  of  which  are  equal.  W^hy  ? 

Z  A'B'D'C     3^ 
4* 


5. 


ZABDC 
Z  A'B'D'C      Z  A'B'C 


Why 


Z ABDC        Z ABC 

Case  II.    When  the  plane  angles  are  incommensurable. 

Hypothesis.     Z  ABC   and   Z  A'B'C,   plane    A   of    diedral 
A  ABDC  and  A'B'D'C,  respectively,  are  incommensurable. 


Conclusion. 


SUPPLEMENTARY  THEOREMS 
Z  A'B'D'C     Z  A'B'C 


345 


Proof.  1.  Let  Z  x\BC  be  divided  into  any  number  of  equal 
parts,  and  let  one  of  these  parts  be  applied  to  Z  A'B'C  as  unit 
of  measure. 

Since  Z  ABC  and  Z  A'B'O'  are  incommensurable,  the  unit 
will  not  be  contained  exactly  in  Z  A'B'C 

A  certain  number  of  equal  parts  will  extend  from  A'B'  to 
B'E,  leaving  the  remainder  Z  EB'C  less  than  the  unit  of 
measure. 

2.   Pass  a  plane  through  B'D'  and  B'E.     Then 

Z  A'B'D'E     Z  A'B'E 


ZABDC       A  ABC 


Case  I 


3.  Now  let  the  number  of  subdivisions  of  Z  ABC  be  in- 
definitely increased ;  then  the  unit  of  measure  will  be  indefi- 
nitely decreased,  and  consequently  the  remainder  Z  EB'C  will 
approach  the  limit  0.  §  401 

Z  A'B'D'E  .  Z  A'B'D'C 


4. 

6.   Also 
6. 


ZABDC        ZABDC 

["  =,"  means  "  approaches  the  limit."] 

Z  A'B'E  .  Z  A'B'C 


ZABC 
Z  A'B'D'C 


ZABC 
Z  A'B'C 


ZABDC        ZABC 


§  403,  (a) 

§  403,  (a) 
§  403,  (h) 


BOOK   YII 


POLYEDEA 


522.  Surfaces.  So  far  only  plane  surfaces  have  been  con- 
sidered. 

A  Curved  Surface  is  a  surface  no  part  of  which  is  plane. 

The  surface  of  a  spherical  object  is  a  familiar  example  of  a  curved  sur- 
face. 

It  is  evident  that  there  may  be  surfaces 
of  which  part  is  plane  and  part  is  curved ; 
as  surface  S. 

Also  there  are  surfaces  consisting  of  two 
or  more  parts  each  of  which  is  plane ;  as 
surface  T. 

523.  Closed  Surfaces.  Let  a  plane  ABC  intersect  the  facfes 
of  triedral  angle  0-XYZ,  and  consider  the  sur- 
face consisting  of  triangles  OAB,  GAG,  GBG, 
and  ABC,  and  the  portions  of  planes  within 
them.  This  surface  incloses  a  finite  portion 
of  space.  Such  a  surface  is  a  closed  surface. 
A  closed  surface  is  such  that  the  intersection 
of  it  made  by  evei-y  intersecting  plane  is  a 
closed  line. 

524.  A  closed  surface  is  convex  if  the  intersection  with  it  of 
every  intersecting  plane  is  a  convex  closed  line. 

It  will  be  assumed  that  all  closed  surfaces  considered  in  this  text  are 
convex. 

525.  A  Solid  is  the  finite  portion  of  space  inclosed  by  a 
closed  surface.  The  surface  is  called  the  surface  of  the  solid, 
and  is  said  to  bound  the  solid. 

346 


POLYEDRA 


347 


In  the  remaining  part  of  solid  geometry  a  detailed  study  is 
made  of  certain  common  solids. 

526.  A  Polyedron  is  a  solid  bounded  by  portions  of  planes, 
called  the  Faces  of  the  polyedron.  The  faces  intersect  in 
straight  lines,  called  the  Edges  of  the  polyedron.  The  edges 
intersect  in  points,  called  the  Vertices  of  the  polyedron.  The 
straight  line  joining  any  two  vertices  of  the  polyedron  which 
do  not  lie  in  the  same  face  is  a  Diagonal  of  the  polyedron. 

527.  The  least  number  of  planes  that  can  form  a  polyedral 
angle  is  three.  Then  the  least  number  of  planes  that  can  form 
a  polyedron  is  four. 

A  polyedron  of  four  faces  is  a  Tetraedron ;  one  of  six  faces 
is  a  Hexaedron ;  one  of  eight  faces  is  an  Octaedron ;  one  of 
twelve  faces  is  a  Dodecaedron ;  and  one  of  twenty  is  an 
Icosaedron. 

The  cube  is  a  familiar  hexaedron. 


Tetraedron 


Pentaedron 


Hexaedron 


Ex.  1.     Verify  for  a  tetraedron,  a  peDtaedron,  and  a  hexaedron,  the 

foniiula 

F-\-  V=E  +  2, 

where  F  is  the  number  of  faces,  V  is  the  number  of  vertices,  and  E  is  the 
number  of  edges. 

Note.  —  This  theorem  is  due  to  the  mathematician  Leonard  Euler. 

A 


Ex.  2.  If  E,  F,  G,  and  H  are  the  mid-points  of 
edges  BD,  BC,  AC,  and  AD,  respectively,  of  tetraedron 
ABCD,  prove  EFGII  a.  parallelogi'am.  D 


348 


SOLID   GEOMETRY  —  BOOK  VII 


Ex.  3.     If  ABCD  is  a  tetraedron,  the  section  made  by  a  plane  parallel 
to  each  of  the  edges  AB  and  CD  is  a  parallelogram. 

Note.  —  Remember  that  by  §  468  a  plane  can  he  drawn  parallel  to  each 
of  two  straight  lines  in  space. 


Ex.  4.  The  lines  joining,  by  pairs,  the  mid-points 
of  opposite  edges  of  any  tetraedron,  intersect  in  a  com- 
mon point. 


Ex.  5.  In  tetraedron  ABCD,  a  plane  is  drawn 
through  edge  CD  perpendicular  to  AB,  intersecting 
faces  ABC  and  ABD  in  CE  and  ED,  respectively.  If 
the  bisector  of  Z  CED  meets  CD  at  F,  prove 

CF :  DF  =  area  ABC  :  area  ABD. 

Suggestion.  —  Recall  §  270. 


PRISMS  AND  PARALLELOPIPEDS 

528.  A  Prism  is  a  polyedron,  two  of  whose  faces  lie  in 
parallel  planes,  and  whose  remaining 
faces,  in  order,  intersect  in  parallel  lines. 
The  parallel  faces  are  the  Bases;  the 
other  faces  are  the  Lateral  Faces;  the 
edges  which  are  not  sides  of  the  bases  are 
the  Lateral  Edges ;  the  perpendicular  be- 
tween the  bases  is  the  Altitude ;  the  sum 
of  the  areas  of  the  lateral  faces  is  the 
Lateral  Area. 

If  a  plane  is  perpendicular  to  the  lateral  edges  of  a  prism, 
its  intersection  with  the  prism  is  called  a  Right  Section  of  the 
prism. 

529.  The  following  Important  Facts  about  a  Prism  should  be 
proved  by  the  pupil : 

I.    TJie  lateral  faces  of  a  prism  are  inclosed  by  parallelograms. 

Prove  BCHG  is  a  O,  using  §  471.  Fig.  §  528 


PRISMS  AND  PARALLELOPIPEDS 


349 


II.  The  lateral  edges  of  a  prism  are  parallel  and  equal. 

III.  The  bases  of  a  prison  are  inclosed  by  congruent  polygons. 
Suggestion.  —  Recall  §  481. 

IV.  Sections  of  the  lateral  surface  of  a 
prison  made  by  two  parallel  planes  cutting 
all  the  lateral  edges  are  congruent  polygons. 

Suggestion.  —  Let  planes  CF  and  C'F'  he  paral- 
lel. 

Prove  CDEFG  ^  C'D'E'F'G'. 

V.  A  section  of  a  prism  made  by  a 
plane  parallel  to  the  base  is  congruent  to 
the  base. 

530.   Kinds  of  Prisms.     A  prism  is  triangular,  quadrangulary 
etc.,  according  as  its  base  is  triangular,  quadran- 
gular, etc. 

A  Right  Prism  is  a  prism  whose  lateral  edges 
are  perpendicular  to  its  bases. 

An  Oblique  Prism  is  a  prism  whose  lateral 
edges  are  not  perpendicular  to  its  bases. 

A  Regular  Prism  is  a  right  prism  whose  base 
is  inclosed  by  a  regular  polygon. 

A  Truncated  Prism  is  that  portion  of  a  prism 
bounded  by  the  base  and  a  plane  not  parallel  to 
the  base,  cutting  all  the  lateral  edges. 

Ex.  6.     Prove  that  every  pair  of  lateral  edges  of  a  prism  determines  a 
plane  which  is  parallel  to  each  of  the  other  lateral  edges  of  the  prism. 

Ex.  7.    Prove  that  the  lateral  edges  of  a  right  prism  are  equal  to  the 
altitude. 

Ex.  8.    Prove  that  the  lateral  faces  of  a  right  prism  are  inclosed  by 

rectangles.  n'  c 

r~- — yf~~-^H 
Ex.  9.     Prove  that  the  section  of  a  prism  made  by  ^-         ^  ^ 

a  plane  parallel  to  a  lateral  edge  is  inclosed  by  a  par- 
allelogram. 

Suggestion.— Jjet  plane  EF'  be  ||  AA',  a  lateral  edge 
of  prism  AC. 


.-L., 


350  SOLID    GEOMETRY  —  BOOK   VII 

Proposition  I.     Theorem 

531.  The  lateral  area  of  a  prism  equals  the  perimeter 
of  a  right  section  multiplied  by  the  length  of  a  lateral 
edge. 


Hypothesis.     DEFGH  is  a  rt.  section  of  prism  AO. 

P=  perimeter  of  DEFGH',  L  =  length  of  A  A' ; 

S  =  the  lateral  area. 
Conclusion.  S  =  LP. 

Proof.     1.  Area  of  O  AB'  =  Lx  DE.  Why  ? 

2.        Similarly,  area  of  CJ  BC'  =  Lx  EF.  Why  ? 

Complete  the  proof. 

532.  Cor.  TJie  lateral  area  of  a  right  prism  equals  the  perim- 
eter of  the  base  multiplied  by  the  length  of  the  altitude. 

Ex.  10.  Find  the  lateral  area  of  a  regular  hexagonal  prism,  each  side 
of  whose  base  is  3  and  whose  altitude  is  9. 

Ex.  11.  There  are  upon  a  porch  six  columns  having  the  form  of  regu- 
lar octagonal  prisms.  If  the  side  of  the  base  is  4  inches  and  the  altitude 
of  the  column  is  7  feet,  find  the  total  of  the  lateral  areas  of  the  columns. 

533.  The  Volume  of  a  Solid  is  a  number  which  indicates  the 
measure  of  that  solid  in  terms  of  a  unit  of  solid  measure ;  it  is 
the  ratio  of  the  solid  to  the  unit  of  solid  measure. 

534.  Two  solids  are  equal  if  they  have  equal  volumes.  Evi- 
dently two  congruent  solids  are  equal ;  likewise  two  solids 
which  can  be  divided  into  parts  which  are  respectively  congru- 
ent are  equal. 


PRISMS  AND  PARALLELOPIPEDS  351 

Proposition  II.     Theorem 

535.  An  oblique  prism  equals  a  right  prism  which 
has  for  its  base  a  right  section  of  the  oblique  prism,  and 
for  its  altitude  a  lateral  edge  of  the  oblique  prism. 


Hypothesis.  Fir  is  a  right  prism.  Its  base  FKis  a  right 
section  of  oblique  prism  AD'. 

Its  altitude  FF'  =  AA\  a  lateral  edge  of  AD'. 
Conclusion.  FK'  =  AD'. 

Proof.     1.  ABCDE^A'B'C'D'E', 

and  FGHKL  ^  F'O'H'K'L'.  §  529,  IV 

2.  AF=  A'F' ;  BG  =  B'G' ;  CH  =  G'H' ;  etc.    Prove  it. 

3.  Slide  polyedron  AK  upward,  letting  AF,  BG,  etc.,  move 
along  lines  AF',  BG',  etc.,  until  ABODE  coincides  with 
A'B'C'D'E'. 

4.  Then  F,  G,  H,  etc.,  fall  upon  F',  G',  H',  etc.      Step  2 

5.  .-.  FGHKL  will  fall  upon  F'G'H'K'L'. 

6.  .-.  polyedron  AK  ^  polyedron  A'K'. 

7.  .♦.  prism  AD'  =  prism  FK'.  §  534 

536.  Cor.  Two  right  prisms  are  congruent,  and  hence  equal, 
when  they  have  congruent  bases  and  equal  altitudes. 

For  the  bases  can  be  made  to  coincide.  Then  the  lateral  edges  of  one 
will  coincide  with  the  homologous  edges  of  the  other  by  §  530  and  Note, 
§  456.     Then  the  upper  bases  must  coincide. 


352 


SOLID   GEOMETRY  —  BOOK  VII 


537.  A  Parallelepiped  is  a  prism  whose  base  is  inclosed  by 
a  parallelogram.     As  a  consequence,  all  the 
faces  of  a  parallelopiped  are  inclosed  by  par- 
allelograms. 

A  Eight  Parallelopiped  is  a  parallelopiped 
whose  lateral  edges  are  perpendicular  to  its 
bases. 

A  Rectangular  Parallelopiped  is  a  right  par- 
allelopiped whose  base  is  inclosed  by  a  rec- 
tangle. Consequently  all  the  faces  are  inclosed 
by  rectangles.     (See  Ex.  8,  p.  349.) 

A  Cube  is  a  rectangular  parallelopiped  whose  six  faces  are 
inclosed  by  squares. 

Ex.  12.     How  many  faces  of  a  right  parallelopiped  are  rectangles  ? 
Ex.  13.     Is  a  cube  a  prism  ? 

Proposition  III.     Theorem 

538.  The  opposite  lateral  faces  of  a  parallelopiped 
are  congruent  and  parallel. 


Hypothesis.     AC   and    A'C   are    the    bases    of    parallelo- 
piped AC. 

Conclusion.     Eaces  AB'  and  DC  are  congruent  and  II. 

Suggestions.  —  1.    To  prove  AB'  ^  DC ,  prove  them  mutually  equiangular 
and  mutually  equilateral. 

2.  To  prove  AB'  \\DC',  recall  §  481. 

539.   Cor.     Any  pair  of  opposite  faces  of  a  parallelopiped  may 
he  taken  as  its  bases. 


PRISMS   AND  PARALLELOPIPEDS  353 

Proposition  IV.     Theorem 

540.  The  plane  through  tivo  diagonally  opposite 
edges  of  a  parallelopiped  divides  it  into  two  equal  tri- 
angular prisms. 


Hypothesis.  Plane  AC  passes  through,  edges  AA'  and  CC 
of  parallelopiped  A'C. 

Conclusion.     Prism  ^'-^B(7=  prism  ^'-^CZ>. 

Proof.  1.  Let  EFGH  be  a  right  section  of  the  parallelo- 
piped, intersecting  plane  AA'C'C  in  EG. 

2.  EFGH  is  a  O.  Prove  it. 

3.  .-.  AEFG^AEGH.  Why? 

4.  Prism  A'- ABO  =  a  right  prism  with  base  EFG  and 
altitude  AA\ 

Prism  A'-ACD  =  Si  right   prism  with  base  EGH  and 
altitude  AA'.  §  535 

6.  But  these  right  prisms  are  equal.  §  536 

6.  .-.  A'-ABC  =  A'-AGD. 

Ex.  14.  Prove  that  the  diagonals  of  a  rectangular  parallelopiped  are 
equal. 


Ex.  15.  Prove  that  the  diagonals  of  a  parallelo-       /  yz^^f^ 
piped  bisect  each  other.  \/^y'~'^^\^ 


B 
Suggestion.  —  Prove  that  each  of  the  other  diagonals  bisects  BG. 

Note.  —  The  point  Of  intersection  of  the  diagonals  of  a  parallelopiped  is 
called  the  center  of  the  parallelopiped. 


354 


SOLID   GEOMETRY  —  BOOK  VII 


Ex.  16.  Prove  that  any  line  drawn  through  the  center  of  a  parallele- 
piped, terminating  in  a  pair  of  opposite  faces,  is  bisected  at  that  point. 

Ex.  17.  Prove  that  the  line  joining  the  center  of  a  parallelepiped  to 
the  center  of  any  face  is  parallel  to  any  edge  of  the  parallelopiped  which 
intersects  that  face,  and  is  equal  to  one  half  of  it. 

Ex.  18.     Prove  that  the  centers  of  two  opposite  faces  of  a  parallelo- 
piped and  the  center  of  the  parallelopiped  are  collinear. 
Suggestion.  —  Recall  Ex.  17  and  §  90. 

Ex.  19.  If  the  four  diagonals  of  a  quadrangular  prism  pass  through  a 
common  point,  the  prism  is  a  parallelopiped. 

Plan.  —  Prove  BCHE  is  a  O. 

Suggestion.  — DE  and  ^Zf  determine  a  plane  (why?) 
which  intersects  planes  DF  and  BH  in   lines  AD  and    ^ 
HE,  respectively.    Compare  AD  and  HE;  also  compare 
AD  2iX\dBC.    Then  compare  BC  and  HE. 

Ex.  20.  Recall  that  the  diagonals  of  a  square  are  equal,  bisect  each 
other,  and  are  perpendicular  to  each  other. 

What  questions  about  the  diagonals  of  a  cube  do  these  facts  suggest  ? 
Prove  the  answers  to  your  questions. 

541.  The  Dimensions  of  a  rectangular  parallelopiped  are  the 
lengths  of  its  three  edges  which  meet  at  any  vertex. 

The  Volume  of  a  rectangular  parallelopiped  is  readily  deter- 
mined when  the  three  dimensions  are  multiples  of  the  linear 
unit  of  measure. 


Thus,  if  the  dimensions  of  P  are  5  units,  4  units, 
and  3  units,  respectively,  the  solid  can  be  divided  into 
60  unit  cubes.  In  this  case  60,  the  number  which 
expresses  the  area,  is  the  product  of  3,  4,  and  5,  the 
three  dimensions. 


/-/-/--/-/-A 

A-/--/--/--/- 

-A\. 

-l-j-l-i- 

1    1    '    1 

___ — I — , — ) — 

A   1  A 

1     !     1     i 

V 

The    following    three    propositions    prove 
that  this  is  the  correct  formula  for  determining  the  volume  of 
any  rectangular  parallelopiped.     Before  taking  up  these  propo- 
sitions, certain  necessary  new  ideas  are  introduced.* 

*  The  class  may  have  studied  limits  in  Plane  Geometry  (§  401  and  follow- 
ing) .  In  that  case  the  following  two  paragraphs  constitute  a  review  of  those 
paragraphs  of  the  Plane  Geometry. 


PRISMS  AND   PARALLELOPIPEDS  355 

542.  a.  A  Variable  is  a  number  which  assumes  different 
values  during  a  particular  discussion. 

Thus  a  number  x  which  assumes  successively  the  values  1,  ^, 
\j  \y  •••  is  a  decreasing  variable,  assuming  ultimately  values 
which  differ  by  as  little  as  we  please  from  zero. 

A  number  y  which  assumes  successively  the  values  1, 1^,  1|, 
1  J,  •••  is  an  increasing  variable^  assuming  ultimately  values  which 
differ  V)y  as  little  as  we  please  from  2. 

h.  A  Constant  is  a  number  which  has  a  fixed  value  through- 
out a  particular  discussion. 

c.  A  Limit  of  a  Variable  is  a  constant  such  that  the  numeri- 
cal value  of  the  difference  between  the  constant  and  the  varia- 
ble becomes  and  remains  less  than  any  small  positive  number. 
We  say  that  a  variable  approaches  its  limit.  If  a  variable  has  a 
limit,  it  has  only  one  limit.  The  symbol  =  means  "  approaches 
the  limit." 

Thus,  X  above  =  0,  and  y  =  2. 

543.  Limits  Theorems.     It  can  be  proved  that : 

a.  If  a  variable  x  approaches  a  finite  limit  I,  then  ex,  where 
c  is  a  constant,  approaches  the  limit  cl.  §  403,  a 

b.  If  two  variables  are  constantly  equal  and  each  approaches 
a  finite  limit,  then  their  limits  are  equal.  §  403,  b 

c.  If  a  variable  x  approaches  a  finite  limit  a,  and  a  variable 
y  approaches  a  limit  6,  then : 

(x  ±  y)  approaches  the  limit  a±b\ 
xy  approaches  the  limit  a  •  6 ; 

and         -  approaches  the  limit  -,  provided  b  is  not  zero. 
y  & 

Ex.  21.  Find  the  length  of  the  diagonal  of  a  rectangular  parallelo- 
piped  whose  dimensions  are  8,  9,  and  12. 

Ex.  22.  Prove  that  the  square  of  a  diagonal  of  a  rectangular  paral- 
lelopiped  is  equal  to  the  sum  of  the  squares  of  its  edges. 

Ex.  23.  Determine  the  length  of  the  diagonal  of  a  cube  whose  edge  is 
of  length  s. 

Note.  —  Supplementary  Exercises  18-22,  p.  456,  can  be  studied  now. 


356 


SOLID    GEOMETRY  —  BOOK  VII 


Proposition  Y.     Theorem 

544.    TiDO  rectangular  parallelopipeds  having  congru- 
ent bases  have  the  same  ratio  as  their  altitudes. 
Case  I.     When  the  altitudes  are  commensuraUe, 


/>-- 

■A 

Q 

/  1 

/      / 1 

A 

/  A 

-/-7 

/  i- 

B 

/       1 

'A 

y   1 

/  J 

V 

/ 

v 

Hypothesis.     P  and  Q  are  rectangular  parallelopipeds  with 
congruent  bases,  and  commensurable  altitudes  AA!  and  BB'. 

Conclusion.  -  =  4^!- 

P     AA 

Proof.     1.     Let  AC^  a  common  measure  of  AA  and  BB'^  be 

contained  4  times  in  AA  and  3  times  in  BB'. 

BB'  J^ 

"  AA     4  * 

2.  Through  the  points  of  division  of  AA  and  BB'  draw 
planes  ±  AA  and  BB' ,  respectively. 

3.  Then  P  is  divided  into  4,  and  Q  is  divided  into  3,  parts 
which  are  all  congruent.  §  536 

4.  •   5-5. 


P     4 
Q^BB' 
P     AA 


Why? 


Case  II  (Fig.  p.  857). 
cojumensurahle* 


When  the  altitudes  are  in- 


Hypothesis.     P  and  Q  are  rectangular  parallelopipeds  with 
congruent  bases  and  incommensurable  altitudes  AA'  and  BB'. 
*  This  proof  may  be  omitted  if  desired. 


PRISMS  AND  PARALLELOPIPEDS 


357 


Conclusion. 


P     AA' 


A 

P 

A 

/ 

L 

A 

/ 

/ 

A / 

A 

// 

/       1 

/ 

/ 

/ 

Proof.  1.  Divide  AA!  into  any  number  of  equal  parts,  and 
apply  one  of  these  parts  to  BB!  as  unit  of  measure.  Since  A  A 
and  BB'  are  incommensurable,  a  certain  number  of  these  parts 
will  extend  from  B  to  O,  leaving  a  remainder  CB'  less  than  the 
unit  of  measure. 

2.  Draw  plane  CD  J_  BB',  and  let  parallelopiped  BD  be 
denoted  by  Q'. 

Then  %  =  ^.'  Case  I 

P      AA 

[Since  A  A  and  BC  are  commensurable.] 

3.  Let  the  number  of  subdivisions  in  AA!  be  indefinitely  in- 
creased. The  length  of  each  subdivision  will  then  diminish 
indefinitely,  and  hence  CB'  will  approach  the  limit  0.      §  542,  c 


Also 


and 
4. 


Qt  Q 

-^  will  approach  the  limit  -^, 

-  will  approach  the  limit  — —  ■ 
AA 


AA' 


Q^BB' 
P     AA' 


§  543,  a 

§  543,  a 
§  543,  b 


545.  Cor.  If  tivo  rectangular  parallelopipeds  have  two  dimen- 
sions of  one  equal  respectively  to  two  dimensions  of  the  other j  their 
volumes  have  the  same  ratio  as  their  third  dimensions. 

Ex.  24.  Compare  the  volumes  of  two  rooms  having  the  same  floor 
space  if  their  heights  are  8'  6"  and  9'  respectively. 


358 


SOLID   GEOMETRY  —  BOOK  VII 


Proposition  VI.     Theorem 

546.    Two  rectangular  parallelopipeds  having  equal 
altitudes  have  the  same  ratio  as  their  bases. 


p 

Q 

n 

A 

c 

A) 

/\ 

c 

A 

/ 

1 

! 

1 

/ 

} 

C 

X 

J 

Hypothesis.  Eectangular  parallelopipeds  P  and  Q  have  equal 

altitudes  c  and  bases  with  dimensions  a,  b,  and  a',  b',  respec- 
tively. 

n^^ -.!«-: P      Ct   X  b 


Q     a'xb' 

Proof.    1.    Let  i2  be  a 
sions  c,  a',  and  b. 

rectangular  parallelopiped 

with  dimen- 

2. 

Then 

P     a 
R-a" 

§545 

and 

R     b 
Q     b'' 

Why? 

3. 

Then, 

multiplying 

the  equations 
P     ab 
Q     a'b' 

of  step  2, 

Ax. 

5,  §51 

Note.  —  Two  rectangular  parallelopipeds  having  a  dimension  of  one 
equal  to  one  dimension  of  the  other,  have  the  same  ratio  as  the  products 
of  their  other  two  dimensions. 

Ex.  25.  Two  rectangular  parallelopipeds,  with  equal  altitudes,  have 
the  dimensions  of  their  bases  6  and  14,  and  7  and  9,  respectively.  Find 
the  ratio  of  their  volumes. 

Ex.  26.  Compare  each  of  the  following  rectangular  parallelopipeds 
with  each  of  the  others  :  i2,  having  dimensions  5,  7,  and  9 ;  ^,  having 
dimensions  9,  5,  and  4  ;   T,  having  dimensions  4,  0,  and  7. 


PRISMS  AND  PARALLELOPIPEDS 


359 


Proposition  VII.     Theorem 
547.    Two  rectangular  parallelopipeds  have  the  same 
ratio  as  the  products  of  their  three  dimensions. 


F 

R 

A            > 

Q 

/V     /\ 

/\ 

/ 

/\     / 

/\ 

/ 

i 
] — 

/ 

c 

/ 

c 

V 

/ 

A) 

/ 

Hypothesis.     P  and  Q  are  rectangular  parallelopipeds  having 
dimensions  a,  6,  c,  and  a',  6',  c',  respectively. 

Conclusion.  ^=-^- 

q     a'b'c' 

Suggestions.  —  1.  Let  ^  be  a  rectangular  parallelepiped  with  dimensions 
a',  b',  and  c. 

2.   Find  -  and  -  by  §  546  and  §  545.     Then  multiply  -  by   -• 
R  Q  R        Q 

Proposition  VIII.     Theorem 
548.    If  the  unit  of  solid  measure  is  the  cube  ivhose 
edge  is  the  linear  unit,  the  volume  of  a  rectangular  par- 
allelopiped  equals  the  product  of  its  three  dimerisions. 


/\ 

/ 

1 

1 

/ 

c 

7 

A 

/ 

1 
>— 

/i 

Hypothesis,     a,  6,  and  c  are  the  dimensions  of  rectangular 
parallelopiped  P,  and  Q  is  the  unit  of  measure. 
Conclusion.  Volume  of  P  =  a  x  6  x  c. 


Proof.     1. 


Volume  of  P  = 


Complete  the  proof,  applying  §  647. 


533 


360  SOLID    GEOMETRY  —  BOOK  VII 

549.  Cor.  1.  Tlie  volume  of  a  cube  is  equal  to  the  cube  of  its 
edge. 

550.  Cor.  2.     The  volume  of  a  rectangular  parallelopiped  is 

equal  to  the  product  of  its  base  and  altitude. 

Note.  —  Corollaries  1  and  2  are  expressed  in  their  commonly  abbreviated 
form.  Expressed  more  accurately,  the  second  would  be  "  the  volume  of  a 
rectangular  parallelopiped  is  equal  to  the  product  of  the  area  of  the  base 
and  the  length  "of  the  altitude."  The  brief  form  of  statement  will  be 
employed  in  the  remaining  theorems  of  solid  geometry. 

Remember  that  volume  of  a  solid  means  the  number  of  cubic  units  in  it. 
In  all  succeeding  theorems  relating  to  volumes,  it  is  understood  that  the 
unit  of  solid  is  the  cube  whose  edge  is  the  linear  unit,  and  the  unit  of 
surface  the  square  whose  side  is  the  linear  unit. 

Ex.  27.  Find  the  ratio  of  the  volumes  of  two  rectangular  parallele- 
pipeds whose  dimensions  are  8, 12,  and  21,  and  14,  15,  and  24,  respectively. 

Ex.  28.  Find  the  volume  and  the  area  of  the  entire  surface  of  a  cube 
whose  edge  is  4  in. 

Ex.  29,  (a)  If  the  edge  of  a  cube  is  e,  express  by  formulae  the 
total  area,  the  volume,  and  the  length  of  a  diagonal.  (6)  Using  the 
proper  formula,  determine  the  edge  when  the  diagonal  is  12.  (c)  Using 
the  proper  formula,  determine  the  edge  when  the  total  area  is  150  sq.  in. 

Ex.  30.  Find  the  altitude  of  a  rectangular  parallelopiped,  the  dimen- 
sions of  whose  base  are  21  and  30,  equal  to  a  rectangular  parallelopiped 
whose  dimensions  are  27,  28,  and  35. 

Ex.  31.  What  must  be  the  height  of  a  tank  having  the  form  of  a 
rectangular  parallelopiped  whose  base  has  the  dimensions  5  ft.  and  8  ft., 
in  order  that  the  tank  will  contain  1800  gal.  of  water  when  the  water  rises 
to  within  one  foot  of  the  top  ?  (One  cu.  ft.  of  water  =  7^  gal.  approxi- 
mately. ) 

Ex.  32.  How  many  barrels  of  water  will  run  into  a  cistern  during  a 
^  in.  fall  of  rain  from  the  roof  of  a  barn  whose  total  roof  area  is  800  sq.  ft. 
(One  cu.  ft.  of  water  =  7|  gal. ) 

Ex.  33.  (a)  How  many  cubic  yards  of  concrete  are  required  for  the 
foundation  walls  of  a  house  25  ft.  x  35  ft.,  if  the  walls  are  10  in.  thick  and 
are  8  ft.  high  ?  (6)  How  many  bags  of  cement  are  required  if  the  mixture 
contains  4  bags  of  cement  to  one  yard  of  gravel  ? 

Note.  —  Supplementary  Exercises  23-28,  p.  456,  can  be  studied  now. 


PRISMS  AND  PARALLELOPIPEDS 


361 


Proposition  IX.     Theorem 

551.    TJie  volume  of  any  parallelopiped  is  equal  to 
the  product  of  its  hose  and  altitude. 


H3rpothesi8.     ^C  is  an  oblique  parallelopiped. 
Let  the  length  of  altitude  AE  be  H,  the  area  of  base  ABCD 
be  B,  and  the  volume  of  AC  be  V. 
Conclusion.  V=BH. 

Proof.     1.   Extend  edges  AB,  A'B',  D'C,  and  DC. 

On  AB  extended,  take  FG  =  AB. 
Draw  planes  FK'  and  GH'  1.  FG,  forming  right  parallelo- 
piped FH'. 

2.  .-.  prism  FH'  =  prism  AC.  §  535 

3.  Extend  edges  HG,  H'G',  K'F,  and  KF. 

On  HG  extended,  take  NM=HG;  draw  planes  NP  and 
ML'  1.  NM,  forming  rectangular  parallelopiped  LN^. 


4.  .-.  prism  L'N=  prism  FH'. 

5.  .'.  prism  L'N=  prism  AC. 

6.  But  volume  L'N  =  LMNP  x  30f . 

7.  .-.  volume  AC  =  LMNP  x  MM. 

8.  But  the  length  of  MM'  =  H 
and  area  LMNP  =  area  ABCD  =  B. 

9.  .-.  vol.  AC  =  BH. 


Why? 
Why? 


Note,  p.  362 


,  Note.  —  The  student's  understanding  of  this  theorem  will  be  increased 
greatly  if  a  model  of  the  above  figure  is  at  hand. 


362  SOLID   GEOMETRY  —  BOOK  VII 

Note  1.  —  Proof  that  LN'  (step  3,  §  551)  is  a  rectangular  parallelopiped. 

1.  Since  FG  ±  plane  GH',  .-.  plane  LH±  plane  MH'.  §  495 

2.  Since  MM'  ±  MN,  .-.  MM'  ±  plane  LH.  §  496 

3.  .-.  Z  LMM'  =  a  rt.  Z. 

4.  .'.  LM'  is  a  rectangle.  §  141 

5.  .-.  LN'  is  a  rectangular  parallelopiped.  §  537 
Note  2.  —  Proof  that  LMNP  =  ABCD. 

LMNP  =  FGHK,  since  Os  having  equal  bases  and  equal  altitudes 
are  equal.     Similarly  FGHK=  ABCD,  and  .-.  LMNP  =  ABCD. 

Proposition  X.     Theorem 

552.    The  volume  of  a  triangular  prism  is  equal  to 
the  product  of  its  base  and  altitude. 


Hypothesis.     C'-ABC  is  a  triangular  prism. 
Length  of  altitude  AE  =  H ;  area  of  A  ABC  =  B ;  volume 
oi  C'-ABC  =V. 
Conclusion.  F=  BH. 

Suggestions.  —  1.  Consider  the  parallelopiped  D'-ABCD,  having  its  edges 
parallel  to  AB,  BC,  and  BB'  respectively. 

2.  Compare  volume  of  C'-ABC  with  that  of  D'-ABCD. 

3.  Express  the  volume  of  D'-ABCD,  and  then  of  C'-ABC. 

Note.  — At  this  point,  the  pupil  should  memorize  the  following  formulae 
if  they  are  not  already  known. 

1.  Area  of  a  A  =  V  s(s  -  a)(^s -h){s- c),  §  335 
where  the  sides  are  a,  6,  and  c,  and  s  =  |(a  +  &  +  c). 

2.  Area  of  an  equilateral  A  of  side  s  =  ^^-^.  (Ex.  29,  p.  199,  Book  IV.) 

4 

sV3 

3.  Altitude  of  an  equilateral  A  of  side  s  = 


PRISMS  AND  PARALLELOPIPEDS  363 

Ex.  34.     Find  the  volume  of  a  regular  triangular  prism  the  side  of 
whose  base  is  6  and  whose  altitude  is  10. 

Ex.  35.     Derive  a  formula  for  the  volume  of  a  regular  triangular 
prism  the  side  of  whose  base  is  s  and  whose  altitude  is  h. 

Ex.  36.     Find  the  lateral  area  and  volume  of  a  right  triangular  prism, 
having  the  sides  of  its  base  4,  7,  and  9,  respectively,  and  the  altitude  8. 

Suggestion.  —  To  determine  the  area  of  the  base,  use  the  first  formula  in 
the  note  of  §  552. 

Ex.  37.   Prove  that  the  volume  of  a  right  triangular  prism  is  equal  to 
the  product  of  the  area  of  any  face  and  one  half  the  altitude  to  that  face. 

Ex.   38.   The  volume  of  a  triangular  prism  is  QQVE,  and  one  side  of 

its  base  is  8.     Find  its  lateral  area. 


Proposition  XI.     Theorem 

553.    The  volume  of  any  prism  is  equal  to  the  prod- 
uct of  its  base  and  altitude. 


Hjrpothesis.     Let  H  =  the  length  of  altitude  AMj  B  =  the 
area  of  base  FGHJK,  and  V=  the  volume  of  prism  AJ. 
Conclusion.  F=  BH. 

Suggestions.  —  1.  Through  edge  AF  mid  diagonals  FH and  FJ  of  the  base, 
pass  planes  AFHC  and  AFJD  dividing  prism  P  into  triangular  prisms  P\y 
P2,  and  Ps,  whose  base  areas  are  Bi,  B^,  and  B^,  resj)ectively,  and  whose  com- 
mon altitude  is  of  length  H. 

2.  Express  the  volumes  of  Pj,  P^,  and  P^.  Add  the  results  and  simplify, 
thus  obtaining  an  expression  for  the  volume  P. 


364  SOLID   GEOMETRY  —  BOOK  VII 

554.  Cor.  1.  Two  prisms  having  equal  bases  and  equal  alti- 
tudes are  equal. 

Suggestion.  —  Let  prisms  P  and  P'  have  equal  bases  B  and  B'  respec- 
tively, and  equal  altitudes  H  and  H' .    Prove  P  =  P' . 

555.  Cor.  2.  Tico  prisms  haviiig  equal  altitudes  have  the 
same  ratio  as  their  bases. 

556.  Cor.  3.  Two  prisms  having  equal  bases  have  the  same 
ratio  as  their  altitudes. 

557.  Cor.  4.  Two  prisms  have  the  same  ratio  as  the  products 
of  their  bases  and  altitudes. 

Ex.  39.  Find  the  volume  of  a  regular  hexagonal  prism  the  side  of 
whose  base  is  4  in.  and  whose  altitude  is  9  in. 

Ex.  40.  Express  by  formulae  the  total  area  and  the  volume  of  a 
regular  hexagonal  prism  whose  base  edge  is  e  and  whose  height  is  h. 

Ex.  41,  A  contractor  agreed  to  dig  a  cellar  at  35^  per  cubic  yard. 
The  lot  was  located  upon  a  hillside  so  that  the  depth  of  the  cellar  at  the 
back  was  9  ft.  and  in  front  5  ft.  If  the  cellar  was  38  ft.  from  the  front 
to  back,  and  was  25  ft.  wide,  how  much  did  the  contractor  receive  ? 

Ex.  42.  How  many  cubic  yards  of  concrete  are  re- 
quired for  a  retaining  wall  2  ft.  thick  whose  dimensions 
are  indicated  on  the  adjoining  figure  ?  " '      W 

Note.  —  Supplementary  Exercises  29-33,  p.  457,  can  be  studied  now. 

PYRAMIDS 

558.  A  Pyramid  is  a  polyedron  bounded  by  three  or  more 
triangular  faces  wliicli  have  a  common  vertex,  and  one  other 
plane  face,  the  Base,  which  intersects  each  of 

the  triangular  faces. 

The  common  vertex  of  the  triangular  faces  is 
the  Vertex  of  the  pyramid  ;  the  triangular  faces 
are  the  Lateral  Faces ;  the  edges  terminating  at 
the  vertex  are  the  Lateral  Edges  ;  the  sum  of  the 
areas  of  the  lateral  faces  is  the  Lateral  Area; 
the  perpendicular  from  the  vertex  to  the  plane  of  the  base  is 
the  Altitude. 


PYRAMIDS 


365 


559.  A  pyramid  is  called  triangular,  quadrangular,  etc.,  ac- 
cording as  its  base  is  triangular,  quadrangular,  etc. 

A  Regular  Pyramid  is  a  pyramid  whose  base 
is  inclosed  by  a  regular  polygon,  and  whose 
vertex  lies  in  the  perpendicular  to  the  base  at 
the  center  of  the  base. 

A  Truncated  Pyramid  is  the  part  of  a  pyramid 
included  between  its  base  and  a  plane  cutting 
all  the  lateral  edges. 

The  base  of  the  pyramid  and  the  section  of  the  cutting  plane 
are  called  the  bases  of  the  truncated  pyramid. 

560.  A  Frustum  of  a  Pyramid  is  a  truncated 
pyramid  whose  bases  are  parallel. 

The  Altitude  of  a  frustum  is  the  perpendic- 
ular between  the  planes  of  the  bases. 

561.  The  following  important  facts 
about  pyramids  should  be  proved  by  the 
pui)il : 

I.  The  lateral  edges  of  a  regular  pyramid 
are  equal. 

II.  The  lateral  faces  of  a  regular  pyramid 
are  inclosed  by  congruent  isosceles  triangles. 

III.  TJie  lateral  faces  of  a  frustum  of  any  pyramid 
closed  by  trapezoids. 

IV.  The  lateral  faces  of  a  frustum  of  a 
regular  pyramid  are  inclosed  by  congruent 
trapezoids. 

Suggestion.  —  Superpose  A  OAB  on  A  OBC. 
Prove  ABB' A'  ^  BCC'B'. 

V.  TJie  lateral  edges  of  a  frustum  of  a 
regular  pyramid  are  equal. 

Note.  —  It  will  be  assumed  that  the  boundary  of  the  base  of  the 
is  a  convex  polygon. 


366  SOLID   GEOMETRY  —  BOOK  VII 

Proposition^  XII.     Theorem 

562.   If  a  pyramid  is  cut  hy  a  plaiie  parallel  to  the 
base : 

I.  The  lateral  edges  and  the  altitude  are   divided 
proportionally. 

II.  The  section  is  sir)iilar  to  the  base. 


B 


Hypothesis.  Plane  A'C^,  parallel  to  the  base  of  pyramid 
0-ABCD,  intersects  faces  OAB,  OBC,  OGD,  and  ODA  in 
lines  A'B',  B'C,  CD',  and  D'A',  respectively,  and  altitude  OP 
at  P'. 

Conclusion.     I.    M'^  M'=  0^=  OP'. 
OA      OB       on      OP 

II.       A'B'O'D'  ~  ABCD. 

Proof  of  I.     1.     Through  0,  pass  plane  MN  ||  ABCD. 

2  .  OA^^OB^^^qc^qD^^qp^         ..^^ 

"  OA       OB       OC      OD       op' 
Proof  of  n.     1.     Z  A'B'C  =  Z  ABC,  Z  B'CD'  =  Z  BCD, 
etc.     Prove  it.  §  481 

^  A^^OAi     B^C^OB:   ^^^ 

AB       OA'     BC       OB'  ^ 

^  .   A^^B^C^OD^^D^A^^  ^^., 

"   AB       BC        CD       DA  ^  ' 

4.  .'.A'B'C'D'^ABCD.  Why? 


PYRAMIDS 


367 


563.  Cor.  1.  Tfie  area  of  a  section  of  a  pyramid  parallel  to 
the  base  is  to  the  area  of  the  base  as  the  square  of  its  distance 
from  the  vertex  is  to  the  square  of  the  altitude  of  the  pyramid. 


Proof.    1. 


2.   But 


ArQ2L  A' B' C D'      A'B'' 


3. 


Area  ABCD 

AB' 

A'B> 
AB 

OA' 
OA 

OP' 
OP 

Area,  A' B' CD' 

OP'^ 

§344 


Area  ABCD 


OP 


564.  Cor.  2.  If  two  pyramids  have  equal  altitudes  and  equal 
basesi  sections  parallel  to  the  bases  at  equal  distances  from  the 
vertices  are  equal. 


Hypothesis.  Pyramids  0-ABC  and  O'-A'B'C  have  the  common 
altitude  //,  and  equal  bases,  ABC  and  A'B'C. 

DEF  and  D'E'F'  are  sections  of  the  pyramids  parallel  to  the  bases  at 
the  distance  h  from  the  vertices  O  and  0',  respectively. 

Conclusion.  DEF=  D'E'P. 

Area  DEF      h^   _„^  Area.  D' E' F^       h^ 


Proof.     1. 


^  and  Area  D'E'Ff 


§563 


Area  ABC     m  Area  A'B'C 

Complete  the  proof. 

Ex.  43.  Prove  that  the  areas  of  sections  of  a  pyramid  made  by 
planes  parallel  to  the  base  have  the  same  ratio  as  the  squares  of  the  dis- 
tances to  the  planes  from  the  vertex. 

Ex.  44.  The  altitude  of  a  pyramid  is  12  in.,  and  its  base  is  a  square 
9  in.  on  a  side.  "What  is  the  area  of  a  section  parallel  to  the  base, 
whose  distance  from  the  vertex  is  8  in.  ? 

Ex.  45.  What  part  of  the  area  of  the  base  of  a  pyramid  is  the  area  of  a 
section  made  by  a  plane  which  is  parallel  to  the  base  and  bisects  the  altitude  ? 

Note.  —  Supplementary  Exercises  34-37,  p.  457,  can  be  studied  now. 


368 


SOLID    GEOMETRY  —  BOOK   VII 


565.   The  slant  height  of  a  regular  pyramid  is  the  altitude  of 

any  lateral  face.     (See  II,  §  561.) 

The  slant  height  of  a  frustum  of  a  regular  pyramid  is  the 
altitude  of  any  lateral  face.     (See  III,  §  561.) 

Ex.  46.     Prove  that  the  perimeter  of  the  mid-section  of  a  frustum  of 
a  pyramid  is  one  half  the  sum  of  the  perimeters  of  the  bases. 

Ex.  47.  What  is  the  slant  height  of  a  regular  quad- 
rangular pyramid  whose  altitude  is  12  and  the  side  of 
whose  base  is  4  ? 

Suggestions.  —  1.  Let  ABV  be  one  face,  VC  be  the  alti- 
tude, and  (7  the  center  of  the  base. 

2.  Determine  the  length  of  AC,  and  of  DC;  then  of  VD. 

Ex.  48.     What  is  the  slant  height  of  a  regular  hexagonal  pyramid 
whose  altitude  is  10  and  whose  base  edge  is  4  ? 


Proposition  XIII.     Theorem 

566.  The  lateral  area  of  a  regular  pyramid  is  equal 
to  the  perimeter  of  its  base  multiplied  by  one  half  its 
slant  height. 


Hypothesis.     0-ABCDE  is  a  regular  pyramid. 
P  =  perimeter  of  its  base ;  L  =  length  of  its  slant  height 
S=  the  lateral  area. 


Conclusion. 
Proof.   1. 


S=iPL. 


iLx  AB. 


Area  of  A  OAB 

Complete  the  proof. 


Why? 


PYRAMIDS  369 

567.  Oor.  Tlie  lateral  area  of  the  frustum  of  a  regular  pyra- 
mid is  equal  to  one  half  the  sum  of  the  perimeters  of  the  bases 
multiplied  by  the  slant  height. 


Hypothesis.     AD'  is  a  frustum  of  a  regular  pyramid. 
L  =  the  length  of  the  slant  height ;  p  and  P  =  the  perimeters  of  the 
upper  and  lower  bases  respectively  ;  and  S  =  the  lateral  area. 

Conclusion.  S=^L(^p  +  P). 

Proof.     1.     Aresiof  ABB' A' =  ^L(AB  + A' B').  Why? 

Complete  the  proof. 

Ex.  49.    What  is  the  slant  height  of  a  regular  triangular  pyramid 
whose  altitude  is  10  and  whose  base  edge  is  4  ? 
Suggestion.  — Recall  §  172. 

Ex.  50.  Express  the  lateral  area  of  a  regular  pyramid  in  terms  of 
the  length  of  the  slant  height  and  the  perimeter  of  the  section  midway 
between  the  base  and  the  vertex. 

Ex.  51.  Prove  the  lateral  surface  of  any  pyramid  greater  than  its 
base,  when  the  perpendicular  from  the  vertex  to  the  base  falls  within  the 
base. 

Suggestion.  —  From  the  foot  of  the  altitude  draw  lines  to  the  vertices  of 
the  base ;  each  A  formed  has  a  smaller  altitude  than  the  corresponding 
lateral  face. 

Ex.  52.  Determine  the  lateral  area  of  each  of  the  pyramids  in  Exer- 
cises 47  to  49. 

Ex.  53.  In  each  of  the  Exercises  47  to  49  pass  a  plane  parallel  to  the 
base  at  a  distance  of  5  in.  from  the  vertex.  Determine  the  lateral  areas 
of  the  resulting  frustums. 

Ex.  54.  Determine  the  total  areas  of  the  pyramids  of  Exercises  47  to 
49. 

Ex.  55.  The  edges  of  the  bases  of  a  frustum  of  a  regular  square 
pyramid  are  5  in.  and  10  in.  respectively,  and  the  altitude  is  6  in.  Deter- 
mine the  slant  height  and  then  the  lateral  area. 


370 


SOLID   GEOMETRY  —  BOOK   VII 


Pkoposition  XIV.     Theokem 
568.    Two  triangular  pyramids  having  equal  altitudes 


and  equal  bases  are  equal. 


Hypothesis.     0-ABC  and  O'-A'B'C  have  equal  altitudes 
and  equal  bases  ABC  and  A'B'C. 


Conclusion. 


0-ABC  =  O'-A'B'C. 


Proof.  1.  Place  the  pyramids  with  their  bases  in  the  same 
plane,  and  let  H  represent  their  common  altitude. 

Divide  H  into  3  equal  parts. 

Through  the  points  of  division  pass  planes  II  to  the  plane  of 
the  bases,  cutting  0-ABC  in  sections  DEF  and  GHK,  and 
O'-A'B'C  in  sections  D'E'F'  and  G'H'K'. 


.'.  DEF  =  D'E'F' 
and  GHK=  G'H'K'. 


§564 


3.  With  ABC,  DEF,  and  GHK  as  lower  bases,  construct 
prisms  X,  Y,  and  Z  with  their  lateral  edges  equal  and  II  to  AD  ; 
with  D'E'F'  and  G'H'E^  as  upper  bases,  construct  prisms  F' 
and  Z',  with  their  lateral  edges  equal  and  II  to  A'D'. 


4. 


.-.  prism  Y=  prism  F' 
and  prism  Z  =  prism  Z'. 


Why? 


PYRAMIDS  371 

6.  Hence  the  sum  of  the  prisms  circumscribed  about 
0-ABC  exceeds  the  sum  of  the  prisms  inscribed  in  O-A'B'O 
by  prism  X. 

6.  Evidently,  0-ABO  <X+Y+Z, 

and  0-ABC  >  Y  +  Z\ 
Likewise  O'-A'B'C  <X+Y-\-Z  and  >  F'  -h  Z\ 

7.  .-.  0-ABC  and  O'-A'B'C  differ  by  less  than  the  differ- 
ence between  the  sum  of  the  circumscribed  prisms  and  the  sum 
of  the  inscribed  prisms ; 

i.e.  0-ABC  and  O'-A'B'C  differ  by  less  than  the  lower 
prism  X. 

8.  By  increasing  indefinitely  the  number  of  subdivisions 
of  H,  the  volume  of  X  can  be  made  less  than  any  assigned 
number,  however  small. 

9.  Suppose  now  that  the  volume  of  0-ABC  and  0- 
A'B^O  differ  by  any  amount  k. 

Since  X  >  the  difference  between  0-ABC  and  0'-A'B'C\ 
then  X  would  be  >  k. 

10.  But  this  contradicts  step  8. 

11.  .-.  O'-A'B'C  and  0-ABC  cannot  differ  at  all ; 

i.e.  O'-A'B'C  =  0-ABC. 

Note.  —  An  interesting  and  instructive  exercise  at  this  point  is  that  of 
proving  the  equality  of  two  triangles  which  have  equal  bases  and  alti- 
tudes, by  a  proof  like  that  given  for  Proposition  XIV.  In  fact,  it  aids  in 
understanding  Proposition  XIV  if  the  exercise  proposed  is  studied  before 
taking  up  §  668. 

The  two  triangles  are  compared  with  two  sets  of  parallelograms,  one 
inscribed  in  one  triangle,  the  other  circumscribed  about  the  other  triangle. 
The  resulting  figures  are  like  the  triangles  AOB  and  A'O'B'  of  the  figure 
of  §  668. 


372  SOLID   GEOMETRY  —  BOOK  VII 

Proposition  XV.     Theorem 

569.    TJie  volume  of  a  triangular  pyramid  is  equal  to 
one  third  the  product  of  its  base  and  altitude. 


Hypothesis.  11=  the  length  of  the  altitude ;  5=  the  area 
of  the  base  ;  and  V=  the  volume  of  pyramid  0-ABC. 

Conclusion.  V^l  HB. 

Proof.  1.  Let  EOD-ABC  be  the  triangular  prism  having 
base  ABC,  and  its  lateral  edges  equal  and  parallel  to  OB. 

2.  Prism  EOD-ABC  is  composed  of  pyramid  0-ABC  and 
pyramid  0-ACDE. 

3.  Divide  0-ACDE  into  two  triangular  pyramids,  0-ACE 
and  0-CDE  by  passing  a  plane  through  E,  0,  and  C. 

4.  In  pyramids  0-ACE  and  0-ECD : 

The  altitudes  are  common.  Why  ? 

Base  ACE  =  base  ECD.  Why  ? 

.-.  0-ACE  =  0-ECD.  Why  ? 

5.  Pyramid  0-ECD  is  the  same  as  pyramid  C-EOD. 

6.  In  pyramids  0-ABC  and  C-EOD  : 

The  altitudes  are  equal.  Why  ? 

Base  ABC  =  base  EOD.  Why  ? 

.-.  0-ABC  =  C-EOD.  Why  ? 

7.  .-.  O-ABC  =  0-ECD  =  0-ACE. 

8.  .-.  0-ABC  =  i  prism  EOD-ABC. 

9.  The  altitude  of  prism  EOD-ABC  =  H,  and  base  =  B. 

10.  .-.  vol.  EOD-ABC  =  HB.  Why  ? 

11.  .-.  vol.  0-ABC  =  i  HB. 


PYRAMIDS 


373 


Ex.  56.  If  the  base  of  a  pyramid  is  a  parallelogram,  the  plane  deter- 
mined by  the  vertex  of  the  pyramid  and  a  diagonal  of  the  base  divides  the 
pyramid  into  two  equal  triangular  pyramids. 

Ex.  57.  Determme  the  ratio  to  a  given  parallelopiped  of  the  pyramid 
whose  lateral  edges  are  the  three  edges  of  the  parallelopiped  which  inter- 
sect at  any  one  corner. 

Ex.  58.  Each  side  of  the  base  of  a  regular  triangular 
pyramid  is  6,  and  its  altitude  is  4.  Find  its  lateral 
edge,  lateral  area,  and  volume. 

Suggestion.  —  In  the  figure,  C  is  the  center  of  the  base. 

Ex  59.  Find  the  area  of  the  entire  surface  and  the 
volume  of  a  triangular  pyramid,  each  of  whose  edges 
is  2. 


Proposition  XVI.     Theorem 
570.    The  volume  of  any  pyramid  is  equal  to  one  third 


the  product  of  its  base  and  altitude. 


B  C 

The  proof  is  like  that  for  §  553. 
Proof  to  be  given  by  the  student. 

571.     Cor.  1.   Any  two  pyramids  having  eqttal  bases  and  equal 
altitudes  are  equal. 

Cor.  2.    Two  pyramids  having  equal  altitudes  have  the  same 
ratio  OS  their  bases. 

Cor.  3.    Two  pyramids  having  equal  bases  have  the  same  ratio 
as  their  altitudes. 

Cor.  4.   Any  two  pyramids  have  the  same  ratio  as  the  products 
of  their  bases  and  altitudes. 


374  SOLID   GEOMETRY  —  BOOK  VII 

Proposition  XYII.     Theorem 

572.  The  volume  of  a  frustum  of  any  pyramid  is 
equal  to  one  third  of  its  altitude  multiplied  hy  the  sum 
of  its  upper  base,  its  lower  base,  and  the  mean  pro- 
portional between  its  bases. 


Hypothesis.  F=the  volume,  B  =  the  area  of  the  lower 
base,  b  =  the  area  of  the  upper  base,  and  H  =  the  length  of 
the  altitude  of  AC,  a  frustum  of  any  pyramid  0-AC. 

Conclusion.  V=iH(B-\-b-{-  VBb). 

Proof.     1.   Draw  altitude  OP,  cutting  A'C  at  Q. 

2.  V=  vol.  0-AC  -  vol.  O-A'C 

=  i^B{H+oq)-\h{oq) 

=  iHB-^iOQ(B-b). 

3.  But  B:b=  OP:  OQ^  §  563 

4.  .-.  VB  :Vb=  OP:  OQ.  Algebra 

5.  .-.  (VB-VI)  :  V6=  (OP  -  OQ)  :OQ  =  H:  OQ. 

§  256 

6.  .-.  OQ  (VB  -  Vb)  =  HVb. 

7.  Multiplying  both  members  by  V-B  -f-  V6, 

...  OQ  {B-b)  =  H(VBb  +  b). 

8.  Substituting  in  step  2  for  OQ(B  —  b)  its  value  from 
step  7,  

V=\HB  +  \H{-\/Bb^+b) 

=  i  H{B  +  ?>  +  V56). 


1 


PYRAMIDS  375 

Ex.  60.  Find  the  volume  of  a  regular  quadrangular  pyramid  each 
side  of  whose  base  is  3,  and  whose  altitude  is  6. 

Ex.  61.  Find  the  volume  of  a  regular  hexagonal  pyramid  each  side  of 
whose  base  is  4,  and  whose  altitude  is  9. 

Ex.  62.  The  slant  height  and  lateral  edge  of  a  regular  quadrangular 
pyramid  are  25  and  \/674,  respectively.     Find  its  lateral  area  and  volume. 

Ex.  63.  Prove  that  the  lines  joining  the  center  of  a  cube  to  the  four 
vertices  of  one  face  are  the  edges  of  a  regular  quadrangular  pyramid  whose 
volume  is  ^  that  of  the  cube. 

Ex.  64.  Express  the  volume  of  a  pyramid  in  terms  of  its  altitude  and 
the  area  of  its  mid-section  parallel  to  the  base. 

Ex.  65.  Find  the  lateral  area  and  volume  of  a  regular  quadrangular 
pyramid,  the  area  of  whose  base  is  100,  and  whose  lateral  edge  is  18. 

Ex.  66.  Find  the  area  of  the  base  of  a  regular  quadrangular  pyramid, 
whose  lateral  faces  are  equilateral  triangles,  and  whose  altitude  is  5. 

Suggestion.  — Represent  the  lateral  edge  and  the  side  of  the  base  by  x. 

Ex.  67.  Find  the  volume  of  a  frustum  of  a  regular  quadrangular 
pyramid,  the  sides  of  whose  bases  are  9  and  5,  respectively,  and  whose 
altitude  is  10. 

■  Ex.  68.  Find  the  volume  of  a  frustum  of  a  regular  triangular  pyra- 
mid, the  sides  of  whose  bases  are  18  and  6,  respectively,  and  whose  alti- 
tude is  24. 

Ex.  69.  Find  the  volume  of  a  frustum  of  a  regular  hexagonal  pyramid, 
the  sides  of  whose  bases  are  8  and  4,  respectively,  and  whose  altitude  is  12. 

Ex.  70.  A  monument  is  in  the  form  of  a  frustum  of  a  regular  quad- 
rangular pyramid  8  ft.  in  height,  the  sides  of  whose  bases  are  3  ft.  and 
2  ft.,  respectively,  surmounted  by  a  regular  quadrangular  pyramid  2  ft. 
in  height,  each  side  of  whose  base  is  2  ft.  What  is  its  weight,  at  180  lb. 
to  the  cubic  foot  ? 

Ex.  71.  The  areas  of  the  bases  of  a  frustum  of  a  pyramid  are  12  and 
75  respectively,  and  its  altitude  is  9.     What  is  the  altitude  of  the  pyramid  ? 

Suggestion.  —  Let  the  altitude  of  the  pyramid  D' ^C ' 

=  z ;  then  a;  —  9  is  the  1  from  its  vertex  to  the  /;      ^1__^/A 

upper  base  of  the  frustum ;  then  use  §  564.  f^Y      '  /^]   ^c\     \ 

Ex.72.     The  lateral  edge  of  a  frustum  of  /     Dil..|__A.l.V--Ac 

a  regular  hexagonal  pyramid  is  10,  and  the  /    /  j\    [     \[  \/ 

sides  of  its  bases  are  10  and  4,  respectively.  1/       V/C^     \/^ 
Find  its  lateral  area  and  volume.                          A  p  g 

Note. — Supplementary  Exercises  38-42,  p.  457,  can  be  studied  now. 


376 


SOLID   GEOMETRY  —  BOOK   VII 


SUPPLEMENTARY  TOPICS 

Five  groups  of  supplementary  material  follow.  None  of  this 
material  is  needed  for  subsequent  parts  of  solid  geometry. 
Each  group  is  independent  of  each  of  the  others.  The  groups 
are  arranged  in  order  of  importance  and  of  interest.  The 
teacher  should  select  from  this  material  such  parts  as  seem 
best  to  meet  the  needs  of  the  class. 

Group  A.     Prismatoids 

573.  A  Prismatoid  is  a  polyedron  bounded  by  two  parallel 
faces  called  Bases,  and  by  a  number  of  lateral  faces  which  are 
bounded  by  either  triangles,  trapezoids,  or  parallelograms. 

The  Altitude  of  a  prismatoid  is  the  perpendicular  between 
the  bases. 

The  Mid-section  of  a  prismatoid  is  the  section  of  the  plane 
parallel  to  the  base  and  midway  between  them. 

Proposition  XVIII.     Theorem 

574.  .  If  the  areas  of  the  lower  and  upper  bases  of  a  prisma- 
toid are  B  and  b,  respectively,  the  area  of  the  mid-section  is  m, 
the  length  of  the  altitude  is  H,  and  the  volume  is  V,  then, 

V=\H{B^b-\-4.m). 


Proof.     1.    Through  any  point  Fof  the  mid-section  and  each 
edge  of  the  prismatoid,  pass  planes.     These  planes  divide  the 


PRISMATOIDS  377 

prismatoid  into  pyramid  V-ABC,  pyramid  V-DEFO,  pyramids 
like  V-BCFy  and  polyedra  like  V-ABED. 

(a)  Volume  V-ABC  =^\'\ H   h  =  \ Ub. 

(b)  Similarly,  volume  V-DEFG  =  |  IIB. 

(c)  To  compute  the  volume  of  V-BCF : 

1.  Draw  VE  and  VS^  thus  forming  A  VRSy  which  is  a  part 
of  the  mid-section. 

2.  Draw  plane  VBS. 

3.  V-BCF  =  V-RSF  +  V-BRS  +  V-BCS. 

4.  V-RSF=\-^H'AVRS  =  \H'AVR1S.        Prove  it. 

5.  V-BRS  =  \-^H'AVRS  =  \H'AVRS.         Prove  it. 

6.  V-BCS=^2 .  V-BRS,  for  A  BGS=2  •  A  BRS.  §  571,  Cor.  2. 

.-.  F-J5C>S'=f  ^.AFie.S'. 

7.  .-.  V-BCF  =iH'A  VRS. 

Similarly,  the  volume  of  any  triangular  pyramid  with  vertex 
V  and  as  base  a  triangular  lateral  face  is  equal  to  |^  ^  multi- 
plied by  that  part  of  m  which  is  in  the  triangular  pyramid. 

(d)  V-ABED  can  be  divided  into  two  triangular  pyramids 
by  passing  a  plane  through  V,  A,  and  E.  Hence  its  volume 
can  be  obtained  as  in  part  (c). 

(c)  Hence,  the  sum  of  all  pyramids  with  vertex  F,  whose 
bases  are  lateral  faces  of  the  prismatoid,  is  ^  H^m. 

(/)  .-.  volume  of  the  prismatoid  =  ^  HB  -f  ^  Hb  +  J  Hm 

=  iH(B-^b-\-4.m). 

Note. — This  Proposition  is  particularly  interesting  not  alone  because 
it  enables  us  to  determine  the  volumes  of  many  irregularly  shaped  figures, 
but  because  it  includes  many  previous  propositions  as  special  cases. 

Ex.  73.  Is  a  prism  a  special  case  of  a  prismatoid  ?  In  a  prism,  what 
relation  is  there  between  J5,  6,  and  m  ?  Does  the  formula  of  §  674  reduce 
to  the  usual  formula  for  the  volume  of  a  prism  ? 

Ex.  74.    Answer  the  same  questioiw  for  a  pyramid. 


378  SOLID   GEOMETRY  —  BOOK  VII 

Group  B.     Truncated  Prisms 
Proposition  XIX.     Theorem 

575.  A  truncated  triangular  prism  is  equal  to  the  sum  of 
three  pyramids  having  as  common  base  the  lower  base  of  the 
given  prism,  and  having  as  their  vertices  the  three  vertices  of  the 
upper  base  of  the  truncated  prism.        p 


Hypothesis.     DEF-ABG  is  a  truncated  triangular  prism. 

Conclusion.     DEF-ABG  =  E-ABC  +  D-ABC  +  F-ABO. 

Proof.  1.  Pass  planes  through  A,  E,  and  O,  and  through 
D,  E,  and  C,  thus  dividing  DEF-ABQ  into  E-ABC,  E- 
ADC,  and  E-DFC. 

2.  E-ABC  is  one  of  the  required  pyramids. 

3.  E-DAC=  B-DAC.  Cor.  1,  §  571 
[For  the  altitudes  from  B  and  E  to  ADC  are  equal.] 

-Bvit  B-DAC  =  D-ABC 
.-.  E-DAC  =  D-ABC,  the  second  required  pyramid. 

4.  E-DFC=B-AFC.  §571 

[For  A  DFC  =  A  AFC,  and  the  altitudes  from  E  and  B  to  DFC  and 
AFC,  respectively,  are  equal.] 

But  B-AFC  =  F-ABC 
.'.  E-DFC  =  F-ABC,  the  third  required  pyramid. 

5.  .-.  DEF-ABC  =  E-ABC  +  D-ABC  +  F-ABC. 

576.  Cor.  1.  The  volume  of  a  truncated  right  triangular 
prism  is  equal  to  one  third  the  base  multiplied  by  the  sum  of  the 
lateral  edges. 


TRUNCATED  PRISMS 


379 


577.  Cor.  2.  Tlie  volume  of  any  tnincated  triangular  prism  is 
equal  to  the  product  of  one  third  the  area  of  a  ^f 

right  section  by  the  sinn  of  the  lateral  edges.  jy^ 

Suggestions.  —  1.  Let  XYZ  be  the  right  section  whose 
area  is  r. 

2.  Apply  §  676,  to  DEF-XYZ,  and  to  ABC-XYZ. 

3.  Prove  DEF-ABC  =  l^r  {AD -\- BE  +  CF) . 

Ex.  75.  Find  the  volume  of  a  truncated  right  triangular  prism,  the 
sides  of  whose  base  are  5,  12,  and  13,  and  whose  lateral  edges  are  3,  7, 
and  5,  respectively. 

Ex.  76.  Find  the  volume  of  a  truncated  right  triangular  prism 
whose  lateral  edges  are  11,  14,  and  17^  having  for  its  base  an  isosceles 
triangle  whose  sides  are  10,  13,  and  13,  respectively. 

Ex.  77.  Find  the  volume  of  a  truncated  regular  quadrangular  prism, 
a  side  of  whose  base  is  8,  and  whose  lateral  edges,  taken  in  order,  are  2, 
C,  8,  and  4,  respectively. 

Suggestion. — Pass  a  plane  through  two  diagonally  opposite  lateral  edges, 
dividing  the  solid  into  two  truncated  right  triangular  prisms. 

Ex.  78.  If  ABCD  is  a  rectangle,  and  EF  is  any 
line  not  in  its  plane  parallel  to  AB,  the  volume  of  the 
solid  bounded  by  figures  ABCD,  ABFE,  CDEF,  ADE, 
and  BCF,  is 

Ih  X  AD  x(2AB-\-  EF), 

where  h  is  the  perpendicular  from  any  point  of  EF  to 
ABCD.  §  577 

Ex.  79.  If  ABCD  and  EFQH  are  rectangles  lying  in  parallel 
planes,  AB  and  BC  being  parallel  to  EF  and  FQ,  respectively,  the  solid 
bounded  by  the  figures  ABCD,  EFGH,  ABFE, 
BCGF,  CDHQ,  and  DAEH,  is  called  a  rec- 
tangular prismoid. 

ABCD  and  EFGHa.Te  called  the  bases  of  the 
rectangular  prismoid,  and  the  perpendicular  dis- 
tance between  them,  the  altitude. 

Prove  the  volume  of  a  rectangular  prismoid 
equal  to  the  sum  of  its  bases,  plus  four  times      A  B 

a  section  equidistant  from  the  bases,  multiplied  by  one  sixth  the  altitude. 

Suggestion.  —Pass  a  plane  through  CD  and  EF,  and  find  the  volumes  of 
the  solids  ABCD-EF  and  EFGH-CD  by  Ex.  78. 

Note.  —  Supplementary  Exercises  43-47,  p.  468,  can  be  studied  now. 


— 

f 

e/- 

1 

<F 

\ 

h' 

-'"n" 

\ 

^ 

kA 

\ 

// 

V- 

380 


SOLID   GEOMETRY  —  BOOK   VII 


Group  C.     Miscellaneous  Theorems 

The  following  theorems  about  tetraedra  are  very  much 
like  certain  theorems  about  triangles  and  about  triedral  angles. 

Proposition  XX.     Theorem 

578.  Two  tetraedra  having  a  triedral  angle  of  one  equal  to 
a  triedral  angle  of  the  other,  have  the  same  ratio  as  the  products 
of  the  edges  including  the  equal  triedral  angles. 

Hypothesis.     V  and    F'  are 

volumes  of  tetraedra  0-ABC 
and  O-A'B'C,  respectively, 
having  the  common  triedral 
ZO. 

Conclusion. 

V  ^   OAx  OBx  00 

V  OA'  X  OB'  X  00'' 

Proof.     1.   Draw  lines  OP  and  O'P'  ±  to  face  OA'B'. 

2.  Let  their  plane  intersect  face  OA'B'  in  line  OPP'. 

3.  Now,  OAB  and  OA'B'  are  the  bases,  and  OP  and  O'P' 
the  altitudes,  of  triangular  pyramids  0-0 AB  and  0'-OA'B\ 
respectively. 

4  .     F^    area  OAB  x  OP 

"  V 


area  OA'B'  x  O'P' 


But 


_  area  OAB 
"  area  OA'B' 
area  OAB 


X 


OP 
O'P'' 
OA  X  OB 


area  OA!B'     OA  x  OB' 


Also  A  OOP  and  00' P'  are  rt.  A. 
Then  A  OOP  and  00' P'  are  similar. 

' '  O'P'      00' ' 
Substituting  from  steps  5  and  8  in  step  4, 


Why? 

(1) 

§346 

Why? 
Why? 

Why? 


V 


OAx  OB       00       OAx  OB  X  00 


OA'  X  OB'      00'      OA'  X  OB'  x  00' 


MISCELLANEOUS  THEOREMS 


381 


Ex.  80.  State  the  theorem  of  plane  geometry  about  triangles  which 
corresponds  to  Proposition  XX. 

Note.  —  For  each  of  the  following  exercises,  state  also  the  correspond- 
ing theorem  about  triangles. 

Ex.  81.  Prove  that  two  tetraedra  are  congruent  if  a  diedral  angle 
and  the  adjacent  faces  of  one  are  congruent,  respectively,  to  a  diedral 
angle  and  the  adjacent  faces  of  the  other,  if  the  congruent  parts  are 
arranged  in  the  same  order. 

Suf/gestion. — Prove  by  superposition. 

Ex.  82.  Two  tetraedra  are  congruent  if  three  faces  of  one  are  con- 
gruent, respectively,  to  three  faces  of  the  other,  if  the  congruent  parts  are 
arranged  in  the  same  order. 

Suggestion.  —  Recall  §  515. 

Ex.  83.  Prove  that  the  three  planes  passing 
through  the  lateral  edges  of  a  triangular  pyramid, 
bisecting  the  sides  of  the  base,  meet  in  a  common 
straight  line. 

Ex.  84.  Prove  that  the  six  planes  through  the 
edges  of  a  tetraedron  bisecting  the  opposite  edges 
meet  in  a  common  point. 

Suggestion.  —  By  Ex.  83,  three  planes  meet  in  line 
VO.  Let  plane  XBC  intersect  VO  at  Y.  Prove  Y  lies 
in  the  remaining  two  planes. 

Note. — The  common  point  is  the  center  of  gravity 
of  the  tetraedron. 

Ex.  85.  Prove  that  the  center  of  gravity  of  a 
tetraedron  divides  the  line  drawn  from  any  vertex 
to  the  center  of  gravity  of  the  opposite  face  in  the  ratio  3:1. 

Ex.  86.  Prove  that  the  six  planes  bisecting  the  diedral  angles  of  a 
tetraedron  meet  in  a  common  point. 

Note. — Pupils  will  find  it  interesting  to  attempt  to  make  up  other 
theorems  about  tetraedra  which  are  suggested  by  theorems  about 
triangles. 

Group  D.  Regular  Polyedra 

679.  A  Regular  Polyedron  is  a  polyedron  whose  faces  are 
congruent  regular  polygons,  and  whose  polyedral  angles  are  all 
congruent. 


382  SOLID   GEOMETRY  — BOOK   VII 

Proposition  XXI.     Theorem 

580.   Not  more  than  Jive  regular  convex  poly edr a  are  possible. 
A  convex  polyedral  Z.  must  have  at  least  three  faces,  and  the 
sum  of  its  face  A  must  be  <  360°.  §  514 

1.  With  equilateral  triangles. 

Since  each  Z  of  an  equilateral  A  is  60°,  we  may  form  a  con- 
vex polyedral  Z.  by  combining  either  3,  4,  or  5  equilateral  A- 

Not  more  than  5  equilateral  A  can  be  combined  to  form  a 
convex  polyedral  Z.  §  514 

Hence  not  more  than  three  regular  convex  polyedra  can 
be  bounded  by  equilateral  A. 

2.  With  squares. 

Since  each  Z  of  a  square  is  90°,  we  may  form  a  convex  poly- 
edral Z  by  combining  3  squares. 

Not  more  than  3  squares  can  be  combined  to  form  a  convex 
polyedral  Z. 

Hence  not  more  than  one  regular  convex  polyedron  can  be 
bounded  by  squares. 

3.  With  regular  pentagons. 

Since  each  Z  of  a  regular  pentagon  is  108°,  we  may  form  a 
convex  polyedral  Z  by  combining  3  regular  pentagons. 

Not  more  than  3  regular  pentagons  can  be  combined  to  form 
a  convex  polyedral  Z.  Why  ? 

Hence  not  more  than  one  regular  convex  polyedron  can  be 
bounded  by  regular  pentagons. 

4.  With  other  regular  polygons. 

Since  each  Z  of  a  regular  hexagon  is  120°,  no  convex  poly- 
edral Z  can  be  formed  by  combining  regular  hexagons.    Why  ? 

Hence  no  regular  convex  polyedron  can  be  bounded  by  reg- 
ular hexagons. 

In  like  manner,  no  regular  convex  polyedron  can  be  bounded 
by  regular  polygons  of  more  than  six  sides. 

Therefore,  not  more  than  five  regular  convex  polyedra  are 
possible. 


REGULAR  POLYEDRA 


383 


Ex.  87.  Prove  that  the  following  construction  produces  a  regular 
tetraedron  having  a  given  side. 

Construction.  —  1.   With  given   side  AB,   construct 
equilateral  ABC. 

2.  At  its  center  E,  draw  ED  1  ABC. 

3.  Take  D  on  DE  so  that  AD  =  AB,  and  draw  AD, 
BD,  and  CD. 

Statement!  —  AB  CD  is  a  regular  tetraedron. 
Prove  its  faces  are  inclosed  by  congruent  equilateral 
As  and  that  its  triedral  angles  are  congruent. 

581.  Models  of  the  five  regular  polyedra  may  be  made  by 
drawing  upon  cardboard  figures  like  the  following. 

Cut  out  the  figures  along  the  outer  line ;  cut  only  halfway 
through  the  cardboard  on  the  inner  lines;  bring  the  edges 
together  and  fasten  them  with  gummed  paper. 


Tbtbaedbon 


Hexaedron 


OCTAEDEON 


DODECAEDBON 


ICOSAEDBON 


384 


SOLID   GEOMETRY  —  BOOK   VII 


Ex.  88.     Make  a  model  of  at  least  one  of  the  regular  polyedra. 

Ex.  89.     What  is  the  sum  of  the  face  angles  at  any  vertex  of  each  of 
the  regular  polyedra  ? 

Ex.  90.     Find  the  volume  and  the  total  area  of  a  regular  tetraedron 
whose  edge  is  10. 

Ex.  91.     Find  the  volume  and  the  total  area  of  a  regular  tetraedron 
whose  edge  is  a. 

Ex.  92.  The  sum  of  the  perpendiculars 
drawn  to  the  faces  from  any  point  within  a 
regular  tetraedron  is  equal  to  its  altitude. 

Suggestion. — Divide  the  tetraedron  into  trian- 
gular pyramids,  having  the  given  point  for  their 
common  vertex.    Find  the  volume  of  each  and  of    b 
the  whole  pyramid  and  form  an  equation. 

Ex.  93.  Prove  that  the  volume  of  a  regu- 
lar octaedron  is  equal  to  the  cube  of  its  edge 
multiplied  by  |Vii. 


Group  E.     Similar  Polyedra 

582.  Two  polyedra  are  similar  when  they  have  the  same 
number  of  faces  similar  each  to  each  and  similarly  placed,  and 
have  their  homologous  polyedral  angles  congruent. 


Ex.  94.  Prove  that  the  ratio  of  any  two  homologous  edges  of  two 
similar  polyedra  is  equal  to  the  ratio  of  any  other  two  homologous 
edges. 

Ex.  95.  Prove  that  any  two  homologous  faces  of  two  similar  polye- 
dra are  to  each  other  as  the  squares  of  any  two  homologous  edges. 

Ex.  96.  Prove  that  the  entire  surfaces  of  two  similar  polyedra 
are  to  each  other  as  the  squares  of  any  two  homologous  edges. 


SIMILAR  POLYEDRA  385 

Proposition  XXII.     Theorem 

583.  Two  tetraedra  are  similar  when  three  face  triangles 
including  a  triedral  angle  of  one  are  similar,  respectively,  to 
three  face  triangles  including  a  triedral  angle  of  the  other,  and 
similarly  placed. 


Hypothesis.     In  tetraedra  ABCD  and  A'B'C'D' 

A  ABC  -  A  A'B'C,  A  ACD  -  A  A' CD',  and 
AADB'^AA'D'B'. 

Conclusion.  ABCD  ~  A'B'C'D. 

Proof.     1.   From  the  given  similar  A,  we  have 

BC     fAC\      CD      fAD\BD  ™     . 

B'C     \A'G')     CD'     \A'D'J     B'D'  ^  ' 

2.  Hence,  A  BCD  -  A  B',C'D'.  Why  ? 

3.  Again,  A  BAC,  CAD,  and  DAB  are  equal,  respectively, 

to  A  B'A'C,  C'A'D',  and  D'A'B'.  Why  ? 

4.  Then,  triedral  A  A-BCD  and  A'-B'C'D'  are  congruent. 

§516 
6.   Similarly,  any  two  homologous  triedral  A  are  congruent. 

6.   Therefore,  ABCD  and  A'B'C'D'  are  similar.  §  582 

Ex.  97.  Two  tetraedra  are  similar  when  a  diedral  angle  of  one  is 
congruent  to  a  diedral  angle  of  the  other,  and  the  face  triangles  including 
the  congruent  diedral  angles  are  similar  each  to  each,  and  similarly  placed. 

Ex.  98.  If  a  tetraedron  be  cut  by  a  plane  parallel  to  one  of  its  faces, 
the  tetraedron  cut  off  is  similar  to  the  given  tetraedron. 


386 


SOLID    GEOMETRY  —  BOOK   VII 


Proposition  XXIII.     Theorem 

584.    Two  similar  tetraedra  have  the  same  ratio  as  the  cubes 
of  any  two  homologous  edges. 


Hjrpothesis.  Fand  V^  are  the  volumes  of  similar  tetraedra 
ABCD  and  A'B'C'D',  respectively,  A  and  A'  being  homologous 
vertices. 


Conclusion. 
Proof.    1. 

2. 
3. 


V^  AB 

Since  ABCD  ~  A'B'C'D', 
triedral  Z.A  =  triedral  Z  A'. 
AB  xACx  AD 


V 


A'B'  X  A'C  X  A'D' 


V     A'B' 


AB  ^  AC  ^^  AD 


But 
and 


A'C"  A'D' 
AC       AB 


A'C     A'B' 
AD       AB 


A'D'     A'B' 
5.   Substituting  from  step  4  in  step  3, 

V      AB  ^  AB  ^  AB 


6. 


A'B'     A'B'     A'B' 

V     aS" 

A^'' 


§578 


Why? 


SIMILAR  POLYEDRA  387 

Note  1.  It  can  be  proved  that  any  two  similar  polyedra  can  be 
divided  into  the  same  number  of  tetraedra,  similar  each  to  each  and  sim- 
ilarly placed.  Consequently,  it  can  be  proved  that  any  two  similar 
polyedra  have  the  same  ratio  as  the  cubes  of  any  two  homologous  edges. 

Note  2.    It  is  interesting  to  notice  that  in  similar  figures 

(a)  two  homologous  lines  have  the  same  ratio  as  any  two  homologous 
sides  ; 

(6)  the  areas  of  two  homologous  limited  or  bounded  surfaces  have  the 
same  ratio  as  the  squares  of  any  two  homologous  sides  ; 

(c)  the  volumes  of  two  homologous  solid  parts  have  the  same  ratio  as 
the  cubes  of  any  two  homologous  sides. 

Ex.  99.  The  volume  of  a  pyramid  whose  altitude  is  7  in.  is  686  cu. 
in.    Find  the  volume  of  a  similar  pyramid  whose  altitude  is  12  in. 

Ex.  100.  If  the  volume  of  a  prism  whose  altitude  is  9  ft.  is  171  cu.  ft., 
find  the  altitude  of  a  similar  prism  whose  volume  is  50f  cu.  ft. 

Ex.  101.  Two  bins  of  similar  form  contain,  respectively,  375  and  648 
bushels  of  wheat.  If  the  first  bin  is  3  ft.  9  in.  long,  what  is  the  length  of 
the  second  ? 

Ex.  102.  A  pyramid  whose  altitude  is  10  in.  weighs  24  lb.  At  what 
distance  from  its  vertex  must  it  be  cut  by  a  plane  parallel  to  its  base  so 
that  the  frustum  cut  off  may  weigh  12  lb.  ? 

Ex.  103.  An  edge  of  a  polyedron  is  66,  and  the  homologous  edge  of  a 
similar  polyedron  is  21.  The  area  of  the  entire  surface  of  the  second 
polyedron  is  135,  and  its  volume  is  162.  Find  the  area  of  the  entire  sur- 
face and  the  volume  of  the  first  polyedron. 

Ex.  104.  The  area  of  the  entire  surface  of  a  tetraedron  is  147,  and  its 
volume  is  686.  If  the  area  of  the  entire  surface  of  a  similar  tetraedron  is 
48,  what  is  its  volume  ? 

Suggestion.  —  Let  z  and  y  denote  the  homologous  edges  of  the  tetraedra. 

Ex.  105.  The  area  of  the  entire  surface  of  a  tetraedron  is  75,  and  its 
volume  is  500.  If  the  volume  of  a  similar  tetraedron  is  32,  what  is  the 
area  of  its  entire  surface  ? 

Ex.  106.  The  homologous  edges  of  three  similar  tetraedra  are  3,  4, 
and  6,  respectively.  Find  the  homologous  edge  of  a  similar  tetraedron 
equivalent  to  their  sum. 

Suggestion.  —  Represent  the  edge  by  z. 


BOOK  VIII 


THE   CYLINDER   AND   THE   CONE 


585.  Generating  a  Surface.     If  a  m 

straight  line  I  moves  so  that  it  con- 
stantly intersects  a  straight  line  b 
and  is  constantly  parallel  to  a 
straight  line  a  which  intersects  b,  it 
can  be  proved  that  /  constantly  lies 
in  the  plane  M  determined  by  a  and 
b ;  also  it  can  be  proved  that  every  point  in  M  lies  in  line  I  at 
some  time  during  its  period  of  movement. 

Line  I  is  said  to  generate  the  plane  M. 

By  suitable  movement  of  a  straight  line,  the  straight  line 
can  be  made  to  generate  various  curved  surfaces  as  well  as  a 
plane  surface. 

586.  A  Cylindrical  Surface  is  the  surface  generated  by  a 
moving  straight  line  which  constantly  intersects  a  given  plane 
curve  and  which  is  constantly  parallel 
to  another  given  straight  line  not  in 
the  plane  of  the  curve. 

Thus,  if  AB  moves  so  as  constantly  to  in- 
tersect plane  curve  AD  and  to  be  constantly    -^ 
parallel  to  MN^  not  in  the  plane  of  AB^  then 
AB  generates  the  cylindrical  surface  BD. 

The  moving  line  is  called  the  Generatrix ;  the  curved  line  is 
called  the  Directrix.  Any  position  of  the  generatrix,  as  EF, 
is  called  an  Element  of  the  cylindrical  surface. 

388 


THE   CYLINDER  AND  THE   CONE  389 

Ex.  1.     How  many  elements  does  a  cylindrical  surface  have  ? 
Ex.  2.     Consider  any  two  elements  of  a  cylindrical  surface.     What 
kind  of  lines  are  they  ? 

Ex.  3.  Can  more  than  one  cylindrical  surface  be  generated  by  use  of 
a  given  directrix  ? 

587.  If  the  directrix  is  a  closed  plane 
curve,  the  cylindrical  surface  separates  an 
infinite  part  of  space  from  surrounding 
space. 

The  cylindrical  surfaces  considered  in  this 
text  always  have  closed  convex  directrices. 

588.  A  Cylinder  is  the  solid  (§  525)  bounded  by  a  portion 
of  a  cylindrical  surface  and  by  portions  of  two  ^ — — ,^^ 
parallel  planes  which  intersect  all  the  elements  j^  j 
of  the  surface.                                                                 /^ -^ 

The  portions  of  the  parallel  planes  are  the     /  / 

Bases  of  the  cylinder ;  and  the  portion  of  the    I,.- - -..^  / 

cylindrical  surface  between  the  planes  is  the  Lat-  f  J 

eral  Surface  of  the  cylinder.  

The  Total  Surface  of  a  cylinder  consists  of  its  lateral  surface 
and  its  bases. 

The  perpendicular  between  the  two  parallel  planes  is  the 
Altitude  of  the  cjdinder.  The  segment  of  an  element  of  the 
cylindrical  surface  which  lies  between  the  bases  is  an  Element 
of  the  cylinder. 

A  Section  of  a  cylinder  is  the  part  of  the  cylinder  common 
to  it  and  a  plane  cutting  all  the  elements  of  the  cylinder. 

If  a  section  of  a  cylinder  is  made  by  a  plane  perpendicular 
to  the  elements,  it  is  a  Right  Section  of  the  cylinder. 

Ex.  4.    The  elements  of  a  cylinder  are  equal  and  parallel. 

589.  Kinds  of  Cylinders.  A  Right  Cylinder  is  a  cylinder 
whose  elements  are  perpendicular  to  its  bases. 

A  Circular  Cylinder  is  a  cylinder  whose  base  is  inclosed  by  a 
circle. 


390  SOLID   GEOMETRY  —  BOOK   VIII 

Proposition  I.     Theorem 

590.  If  a  plane  jJCtsses  through  an  element  of  a  cylin- 
der and  through  at  least  one  other  point  of  the  surface 
of  the  cylinder,  the  intersection  ivith  the  total  surface 
of  the  cylinder  is  a  parallelogram. 


Hypothesis.  Plane  M,  passing  through  element  AB  of  cylin- 
der AF,  intersects  the  lateral  surface  of  ^i^in  point  C,  not  in  JLB. 

Conclusion.  The  section  of  AF  made  by  plane  M  is  inclosed 
by  a  parallelogram. 

Proof.     1.  Through  C  draw  CD  II  AB, 

2.  .-.  CD  is  an  element  of  surface  AF,  intersecting  the  bases 
at  C  and  D  respectively.  §  586 

3.  CD  is  also  in  M.  §  447,  I  and  III 

4.  .-.  CD  is  in  the  intersection  of  AF  and  M. 

5.  BC  and  AD  are  il  straight  lines,  lying  in  both  M  and  the 
bases  of  AF.  Why  ? 

6.  .-.  ABCD  is  the  intersection  of  ^i^and  M. 

7.  Also  ABCD  is  a  O.  Why  ? 

Ex.  5.     If  a  rectangle  revolves  about  one  of  its  sides 


as  an  axis,  it  generates  a  right  circular  cylinder. 

Suggestion. — Prove  that  C  describes  a  circle,  that  BC 
generates  a  cylinder,  and  that  the  cylinder  is  a  right  cylin- 
der. 

591.  As  a  consequence  of  Ex.  5,  a  right  cir- 
cular cylinder  is  also  called  a  cylinder  of  revolution. 


THE   CYLINDER  AND  THE   CONE  391 

Proposition  II.     Theorem 
592.    The  bases  of  a  cylinder  are  congruent 


JE^ ^ 

Hypothesis.     ABf  is  any  cylinder,  with  bases  A'B'  and  AB. 

Conclusion.  Base  A'B'  ^  base  AB. 

Proof.  1.  Let  E'  and  F'  be  two  particular  points  of  curve 
A'B'  and  G'  any  other  point.  Draw  elements  EE'j  FF'j  and 
GG' ;  also  draw  EF,  FG,  EG,  E'F',  F'G',  and  E'G'. 

2.  .-.  A  EFG  ^  A  E'F'G'.  Prove  it. 

3.  .*.  base  AB^  may  be  superposed  on  base  AB  so  that  E\ 
F\  and  &  will  fall  upon  E,  F,  and  G  respectively. 

4.  But  G'  was  any  point  of  AB'  except  E'  and  F\ 

5.  .-.  E\  F,  and  every  other  point  of  curve  AB^  will  fall 
upon  a  corresponding  point  of  curve  AB. 

6.  .-.  base  AB  ^  base  AB.  Why  ? 

Note.  —  It  is  obvious  that  every  point  of  curve  AB  can  be  proved  to 
fall  upon  a  corresponding  point  of  A'B'. 

Ex.  6.  Prove  that  a  section  of  a  circular  cylinder  made  by  a  plane 
parallel  to  the  base  is  inclosed  by  a  circle. 

Note.  — A  section  of  a  circular  cylinder  made  by  a  plane  not  parallel 
to  the  base  is  inclosed  by  a  curve  called  an  ellipse. 

Ex.  7.  Prove  that  a  line  drawn  parallel  to  the 
elements  of  a  circular  cylinder  from  the  center  of  one 
base  intersects  the  other  base  at  its  center. 


Suggestions.  —  1.  Let  A'  be  a  particular  point  and 
B'  be  any  other  point  of  the  bounding  circle  of  the 
upper  base  whose  center  is  C . 

2.  Draw  C'C  II  B'B,  and  draw  element  A' A. 

3.  Prove  that  CB  =  CA,  and  that  C  is  the  center  of  curve  AB. 


392 


SOLID    GEOMETRY  —  BOOK   VIII 


593.    The  Axis  of  a  circular  cylinder  is  a  straight  line  drawn 
between  the  centers  of  its  bases. 

Ex.  8.    Prove  that  the  axis  of  a  circular  cylinder  is  parallel  to  the 
elements  of  the  lateral  surface. 

Ex.  9.     Prove  that  the  axis  of  a  circular  cylinder  passes  through  the 
centers  of  all  sections  parallel  to  the  bases. 


MEASURING  THE  CYLINDER 

594.  In  measuring  the  cylinder,  difficulties  are  encountered 
which  are  like  those  met  in  measuring  a  circle.  For  example, 
the  ratio  of  the  cylindrical  surface  to  the  customary  unit  of 
surface  measure  cannot  have  meaning  in  the  ordinary  sense 
since  they  are  not  magnitudes  of  the  same  kind,  one  being  a 
plane  and  one  a  curved  surface. 

595.  Application  of  Limits  to  the  Circle. 

(a)  The  circumference  of  a  circle,  i.e.  the  length  of  a  circle, 
is  defined  to  be  the  limit  of  the  perimeter  of  any  regular 
inscribed  polygon  as  the  number  of  sides  is  indefinitely 
increased.  §  412 


(h)  It  is  proved  that  the  perimeter  of  any  regular  circum- 
scribed polygon  (as  well  as  inscribed  polygon)  approaches  the 
circumference  of  the  circle  as  limit  if  the  number  of  sides  is 
indefinitely  increased.  §  413 

(c)  It  can  be  proved  that  the  perimeter  of  any  inscribed  or  cir- 
cumscribed polygon  approaches  the  circumference  of  the  circle 
as  limit  if  the  number  of  sides  is  increased  indefinitely  in  such 
manner  that  the  length  of  every  side  approaches  the  limit  zero. 


MEASURING  THE  CYLINDER  393 

(d)  The  area  of  a  circle  is  defined  to  be  the  limit  of  the 
area  of  any  regular  inscribed  polygon  as  the  number  of  sides 
increases  indefinitely.  §  417 

(e)  It  is  proved  that  the  area  of  any  regular  circumscribed 
polygon  approaches  the  area  of  the  circle  as  limit  as  the  num- 
ber of  sides  is  increased  indefinitely.  §  418 

(/)  It  can  be  proved  that  the  area  of  any  inscribed  or  cir- 
cumscribed polygon  approaches  the  area  of  the  circle  as  limit 
if  the  number  of  sides  is  increased  indefinitely  in  such  manner 
that  the  length  of  each  side  approaches  the  limit  zero. 

Note.  — Remember  that  =  is  the  symbol  for  "  approaches  the  limit.'* 

596.  A  prism  is  inscribed  in  a  cylinder  when  its  lateral  edges 
are  elements  of  the  cylinder  and  its  bases  are  ^^^-^•^^ 
in  the  planes  of  the  bases  of  the  cylinder.           J^'^^,'  ^^ 
The   polygons   bounding   the   bases   of    the         J^^^y^^^^^ 
prism  are  inscribed  in  the  boundaries  of  the        /        /  /       / 
bases  of  the  cylinder.                                              /        /  /       / 

597.  Application  of  Limits  to  a  Cylinder.      h-^''t^>J 

Inscribe   in  a  circular   cylinder   a   prism    \x/  ..■';^ 
having  its  base  inclosed  by  a  regular  poly- 
gon ;  then  inscribe  a  second  prism  whose  base  is  inclosed  by 
a  regular  polygon  having  double  the  number  of  sides  ;  imagine 
that  this  process  is  continued  indefinitely.    Pass  a  plane  form- 
ing right  sections  of  the  cylinder  and  the  prisms. 

It  will  be  assumed  evident  that  the  prisms  come  nearer  and 
nearer  to  occupying  the  same  space  as  the  cylinder.  As  a 
consequence : 

(a)  The  Volume  of  the  Cylinder  is  defined  to  be  the  limit  of 
the  volume  of  the  inscribed  prism  as  the  number  of  faces  in- 
creases indefinitely. 

(h)  The  Lateral  Area  of  a  Cylinder  is  defined  to  be  the  limit 
of  the  lateral  area  of  the  inscribed  prism  as  the  number  of 
faces  increases  indefinitely. 

(c)  The  length  of  a  right  section  of  the  lateral  surface  of 
the  cylinder  is  defined  to  be  the  limit  of  the  length  of  the  right 


394  SOLID   GEOMETRY  —  BOOK  VIII 

section  of  the  lateral  surface  of  the  inscribed   prism  as  the 
number  of  faces  increases  indefinitely. 

(d)  It  is  evident  that  the  edge  and  altitude  of  the  inscribed 
prism  equal  respectively  the  element  and  altitude  of  the 
cylinder. 

Proposition  III.     Theorem 

598.  The  lateral  area  of  a  circular  cylinder  is  equal 
to  the  perimeter  of  a  right  section  multiplied  by  the 
length  of  an  element. 


Hypothesis.  8  =  the  lateral  area,  P  =  the  perimeter  of  a 
right  section,  E  =  the  length  of  an  element  of  a  circular 
cylinder. 

Conclusion.  S  =  E  x  F. 

Proof.  1.  Inscribe  in  the  cylinder  a  prism  whose  base  is 
inclosed  by  a  regular  polygon. 

Let  S'  =  the  lateral  area  and  P'  =  the  perimeter  of  a  right 
section. 

2.  Then  S' =  E  x  P'.  Why? 

3.  Let  the  number  of  faces  of  the  prism  increase  indefinitely. 

Then 

S'  =  S,  and  P'  =  P.     §  597,  6  and  c ;  also  Note,  §  595, 

4.  .'.  ExF^ExP.  §  543,  a 

5.  .:  S  =  E  xP.  §  543,  6 

599.  Cor.  1.  The  lateral  area  of  a  right  circular  cylinder  is 
equal  to  the  circumference  of  its  base  multiplied  by  the  length  of 
the  altitude. 


MEASURING  THE   CYLINDER  395 

600.  Cor.  2.  If  R  =  the  length  of  the  radius  of  the  base,  11  = 
the  length  of  the  altitude^  S  =  the  lateral  area,  and  T  =  the  total 
area  of  a  right  circular  cylinder,  then 

(a)  S  =  2irRH. 

(6)   T=27rE^  +  27rRH=27rR(E  +  H). 

Ex.  10.  Find  the  lateral  area  of  a  right  circular  cylinder  whose  alti- 
tude i.s  16  and  the  diameter  of  whose  base  is  18. 

Ex.  11.  Find  the  total  area  of  a  cylinder  of  revolution  whose  altitude 
is  15  and  the  radius  of  whose  base  is  5. 

Ex.  12.  Determine  the  cost  at  15^  per  square  yard  of  painting  the 
vertical  surface  and  top  of  a  gas  holder  whose  diameter  is  30  ft.  and 
wliose  height  is  20  ft. 

Ex.  13.  How  many  square  feet  of  tin  are  required  to  make  30  sec- 
tions of  hot-air  furnace  pipe  10  in.  in  diameter  and  30  in.  in  length  ? 

Proposition  IV.     Theorem 

601.  The  voluine  of  a  circular  cylinder  is  equal  to  the 
area  of  its  base  multiplied  by  the  length  of  its  altitude. 

Hypothesis.  V  =  the  volume,  B  =  the  area  of  the  base,  and 
//  =  the  length  of  the  altitude  of  a  circular  cylinder. 

Conclusion.  V=  H  x  B. 

Proof.  1.  Inscribe  in  the  cylinder  a  prism  having  its  base 
inclosed  by  a  regular  polygon.  Let  F'=  the  volume  and  J5'  = 
the  area  of  the  base. 

2.  .-.  F'  =  //X  i?'.  Why? 

3.  Increase  indefinitely  the  number  of  faces  of  the  prism. 
Then  V^  =  V,  and  2J'  =  B. 

4.  .'.HxB'  =  HxB,  §543,  a. 

5.  .'.V=HxB.  §543,6. 

602.  Cor.  If  V  =  the  volume,  H  =  the  length  of  the  altitude, 
and  R  =  the  radius  of  the  base  of  a  circiUar  cylinder,  then 

F=  wR'H. 


396  SOLID   GEOMETRY  — BOOK  VIII 

Ex.  14.  What  is  the  cost  of  digging  a  dry  well  5  ft.  in  diameter  and 
15  ft.  deep  at  50  f  per  cubic  yard  ? 

Ex.  15.  What  is  the  capacity  in  gallons  of  a  water  tank  12  ft.  in 
length  and  36  in.  in  diameter,  estimating  1\  gal.  of  water  to  a  cubic  foot  ? 

Ex.  16.  How  many  cubic  feet  of  metal  are  there  in  a  hollow  cylin- 
drical  tube  18  ft.  long,  whose  outer  diameter  is  8  in.  and  whose  thickness 
is  1  in.  ? 

Ex.  17.  Determine  the  number  of  cubic  yards  of  concrete  required 
for  the  wall  and  floor  of  a  circular  cistern  8  ft.  in  outside  diameter,  and 
12  ft.  deep,  if  the  walls  and  floor  are  8  in.  thick. 

Ex.  18.  Determine  the  diameter  of  a  2-bbl.  water  reservoir  having 
the  form  of  a  right  circular  cylinder  if  the  length  is  4  ft.  (2  bbl.=63  gal. ; 
1  cu.  ft.  contains  7|  gal.) 

Note.  —  Supplementary  Exercises  48-58,  p.  459,  can  be  studied  now. 

THE  CONE 

603.  A  Conical  Surface  is  the  surface  generated  by  a  moving 
straight  line,  which  constantly  intersects  a  given  plane  curve 
and  constantly  passes  through  a  given  point 
not  in  the  plane  of  the  curve. 

Thus,  if  line  OA  moves  so  as  constantly  to  in- 
tersect plane  curve  ABC,  and  constantly  passes 
through  point  O,  not  in  the  plane  of  the  curve,  it 
generates  a  conical  surface. 

The  moving  line  is  called  the  Generatrix, 
and  the  curve  the  Directrix. 

The  given  point  is  called  the  Vertex,  and  any  position  of  the 
generatrix,  as  OB,  is  called  an  Element  of  the  surface. 

If  the  generatrix  be  supposed  to  be  indefinite  in  length,  it 
will  generate  two  conical  surfaces  of  indefinite  extent,  0- 
A'B'C  and  0-ABC. 

These  are  called  the  upper  and  lower  nappes,  respectively,  of 
the  conical  surface. 

It  will  be  assumed  that  the  directrix  is  a  closed  plane  curve 
so  that  each  nappe  of  the  surface  separates  an  infinite  portion 
of  space  from  surrounding  space. 


THE   CONE  397 

604.  A  Cone  is  a  solid  bounded  by  a  portion  of  one  nappe 
of  a  conical  surface  and  that  part  of  a  plane  cutting  all  the 
elements  of  the  surface  which  lies  within  the 
surface. 

The  plane  is  called  the  Base  of  the  cone,  and  the 
conical  surface  the  Lateral  Surface. 

The  Altitude  of  a  cone  is  the  perpendicular  from 
the  vertex  to  the  plane  of  the  base. 

605.  Kinds  of  Cones. 

A  Circular  Cone  is  a  cone  whose  base  is  inclosed  by  a  circle. 

The  Axis  of  a  circular  cone  is  a  straight  line  drawn  from  the 
vertex  to  the  center  of  the  base. 

A  Right  Circular  Cone  is  a  circular  cone  whose  axis  is  per- 
pendicular to  its  base. 

A  Frustum  of  a  Cone  is  a  portion  of  a  cone  included  between 
the  base  and  a  plane  parallel  to  the  base. 

The  base  of  the  cone  is  called  the  lower  hose,  and  the  section 
made  by  the  plane  the  upper  base,  of  the  frustum. 

The  altitude  of  a  frustum  is  the  perpendicular  between  the 
planes  of  the  bases. 

Ex,  19.     Prove  that  the  elements  of  a  right  circular  cone  are  equal. 

Ex.  20.  Prove  that  the  elements  of  a  frustum  of  a  right  circular  cone 
are  equal. 

Ex.  21.  If  a  right  triangle  be  revolved  about  one  of  its  legs  as  an  axis, 
it  generates  a  right  circular  cone. 

606.  As  a  consequence  of  Ex.  19,  the  distance  from  the  ver- 
tex to  any  point  of  the  circle  bounding  the  base  of  a  right  cir- 
cular cone  is  called  the  Slant  Height  of  the  right  circular  cone. 

As  a  consequence  of  Ex.  20,  the  portion  of  the  slant  height 
of  a  right  circular  cone  between  the  base  and  a  plane  parallel 
to  the  base  is  called  the  Slant  Height  of  the  frustrum  of  the 
right  circular  cone. 

As  a  consequence  of  Ex.  21,  a  right  circular  cone  is  also 
called  a  cone  of  revolution. 


398  SOLID   GEOMETRY  — BOOK  VIII 

Pkoposition  y.     Theorem 

607.  If  a  plane  passes  through  an  element  of  a  cone 
and  through  at  least  one  other  point  of  the  surface  of 
the  cone,  the  intersection  with  the  total  surface  of  the 
cone  is  a  triangle. 


Hypothesis.     Plane  M,  passing  through  element  OC  of  cone 
0-AB,  intersects  the  surface  again  in  point  D,  not  in  OC. 
Conclusion.  Section  OCD  is  a  A. 

Proof.     1.  CD  is  a  straight  line.  Why  ? 

2.  Since  the  base  AB  is  inclosed  by  a  closed  line,  line  CD 
intersects  it  in  two  points  C  and  D  which  lie  in  the  conical 
surface. 

3.  Lines  OC  and  OD  are  elements  of  the  surface.       Def . 

4.  Also  OC  and  OD  lie  in  the  plane  OCD.  Why  ? 

5.  .'.  the  complete  intersection  of  the  surface  and  the  plane 
is  the  triangle  OCD. 

Ex.  22.  Find,  correct  to  one  decimal,  the  slant  height  of  a  right  cir- 
cular cone  whose  altitude  is  9  in.  and  the  radius  of  whose  base  is  3  in. 

Ex.  23.  Find  the  slant  height  of  a  right  circular  cone  whose  altitude 
is  h  and  the  radius  of  wjiose  base  is  r. 

Ex.  24.  The  radii  of  the  upper  and  lower  bases  of  the  frustum  of  a 
right  circular  cone  are  3  in.  and  5  in.  respectively;  the  altitude  of  the 
frustum  is  6  in.     Determine  the  slant  height  correct  to  one  decimal  place. 

Ex.  25.  Find  the  altitude  of  a  right  circular  cone  whose  slant  height 
is  13  and  the  radius  of  whose  base  is  5. 

Ex.  26.  Kepeat  Ex.  24  when  the  radii  are  r  and  B  respectively  and 
the  altitude  is  if. 


THE   CONE 


399 


Proposition  VI.     Theorem 

608.    TJie  section  of  the  lateral  surface  of  a  circular 
cone  made  by  a  plane  2Mrallel  to  the  base  is  a  circle. 


Hypothesis.  A'B'C  is  the  section  of  circular  cone  S-ABC, 
made  by  a  plane  II  base  ABC  whose  center  is  O. 

Conclusion.  A'B'C  is  a  ©. 

Proof.     1.   Draw  axis  O/S,  intersecting  A'B'C  at  O'. 

2.  Let  A'  and  B'  be  any  two  points  in  curve  A'B'C.  Let 
planes  A'O'S  and  B'O'S  intersect  the  base  in  radii  OA  and 
OB,  the  cutting  plane  in  lines  O'A'  and  O'B',  and  the  lateral 
surface  in  lines  SA'A  and  SB'B  respectively. 


3.  SA'A  and  SB'B  are  straight  lines. 

4.  A  SA'O'  ~  A  SAO,  and  A  SB'O'  -  A  SBO. 

5  .    O'^'  ^  O'B' 

"    OA       OB' 

.-.  O'A'  =  O'B'. 


§  607 
Prove  it. 

Prove  it. 
Prove  it. 


6. 

7.  Since  A'  and  B'  are  any  two  points  of  curve  A'B'C  and 
are  equidistant  from  O',  curve  A'B'C  is  a  circle  and  0'  is  its 
center. 

609.  Cor.  The  axis  of  a  circular  cone  passes  through  the 
center  of  every  section  parallel  to  the  base. 

Ex.  27.  Prove  that  the  radii  of  the  upper  and  lower  bases  of  a  frus- 
tutn  of  any  cone  have  the  same  ratio  as  the  distances  of  the  bases  from 
the  vertex  of  the  cone. 


400  SOLID   GEOMETRY  —  BOOK  VIII 

MEASURING  THE  CONE 

610.  A  pyramid  is  inscribed  in  a  cone  when  y 
its  lateral  edges  are  elements  of  the  cone  and  /I 
the  bounding  polygon  of  the  base  of  the  pyra-  /  / 
mid  is  inscribed  in  the  boundary  of  the  base  y^-'-'---y'''''^ 
of  the  cone.  The  vertex  and  altitude  of  the  f  \  /  \  j 
pyramid  coincide  with  the  vertex  and  the  ^"^-^^^U:::!':::^^^^ 
altitude  of  the  cone. 

611.  A  frustum  of  a  pyramid  is  inscribed  in  a  frustum  of  a 
cone  when  its  lateral  edges  are  elements  of 

the  frustum  of  the  cone,  and  the  boundaries  /S^      ^"Tk 

of  the  bases  of  the  frustum  of  the  pyramid  /k\           M  J  . 

are  inscribed  in  the  boundaries  of  the  bases  //    /         /  /  / 

of  the  frustum  of  the  cone.     The  altitude     //-  -/ ,.  /  /  / 

of  the  frustum  of  the  pyramid  coincides  with  /'  i  /             \  k/ 

the  altitude  of  the  frustum  of  the  cone.  \l/               '{/ 

612.  Application  of  limits  to  the  Cone. 

Inscribe  in  a  circular  cone  a  pyramid  having  a  base  inclosed 
by  a  regular  polygon ;  then  inscribe  a  second  pyramid  whose 
base  is  inclosed  by  a  regular  polygon  having  double  the  num- 
ber of  sides  ;  imagine  that  this  process  is  continued  indefinitely. 
It  will  be  assumed  evident  that  the  pyramids  come  nearer 
and  nearer  to  occupying  the  same  space  as  the  cone.  As  a 
consequence : 

(a)  The  Volume  of  a  Cone  is  defined  to  be  the  limit  of  the 
volume  of  a  regular  inscribed  pyramid  as  the  number  of  faces 
is  increased  indefinitely. 

(6)  The  Lateral  Area  of  a  Cone  is  defined  to  be  the  limit  of 
the  lateral  area  of  a  regular  inscribed  pyramid  as  the  number 
of  faces  is  increased  indefinitely. 

(c)  It  can  be  proved  that  the  slant  height  of  a  right  circular 
cone  is  the  limit  of  the  slant  height  of  a  regular  inscribed  pyra- 
mid as  the  number  of  faces  is  increased  indefinitely. 

(d)  The  volume  and  lateral  area  of  a  frustum  of  a  cone  are 


MEASURING  THE   CONE  401 

defined  to  be  the  limits  of  the  volume  and  area  respectively 
of  the  frustum  of  a  regular  inscribed  pyramid,  having  its  base 
inclosed  by  a  regular  polygon,  as  the  number  of  faces  is  in- 
creased indefinitely ;  also,  .the  slant  height  of  a  frustum  of  a 
right  circular  cone  can  be  proved  to  be  the  limit  of  the  slant 
height  of  the  frustum  of  a  regular  inscribed  pyramid  as  the 
number  of  faces  is  increased  indefinitely. 

Proposition  YII.     Theorem 

613.  The  lateral  area  of  a  right  circular  cone  is  equal 
to  the  circumference  of  its  base  multiplied  by  one  half 
its  slant  height. 


Hypothesis.  S  =  the  lateral  area,  C  =  the  circumference  of 
the  base,  and  L  =  the  slant  height  of  a  right  circular  cone. 

Conclusion.  >S  —  ^  CL. 

Proof.  1.  Inscribe  in  the  cone  a  pyramid  whose  base  is 
inclosed  by  a  regular  polygon.  Let  S'  =  the  lateral  area,  P'= 
the  perimeter  of  the  base,  and  L'  =  the  slant  height  of  the 
pyramid. 

2.  Then  S' =  ^  P'lJ.  Why  ? 

3.  Let  the  number  of  faces  of  the  pyramid  increase  indefi- 
nitely, keeping  the  pyramid  always  a  regular  pyramid.     Then 

S'  =  iS,   P'  =  C,   and  V  ^  L.  §  612,  6,  c 

4.  .-.  \P'L'  ^\CL.  §  543,  c 

5.  .'.  S=^\CL  §  542,  6 


402  SOLID   GEOMETRY  —  BOOK  VIII 

614.  Cor.  If  S  denotes  the  lateral  area,  T  the  total  area,  L 
the  slant  height,  and  R  the  radius  of  the  base  of  a  right  circular 
cone,  then  8  =  2  .R^^^  L  =  .RL. 

Also,  T  =  ttRL  +  irR''  =  irR^L  +  R). 

Ex.  28.  Find  the  lateral  area  and  the  total  area  of  a  right  circular 
cone,  the  radius  of  whose  base  is  7  in.  and  whose  slant  height  is  25  in. 

Ex.  29.  How  many  square  yards  of  canvas  are  required  for  a  circus 
tent  having  the  form  of  a  right  circular  cylinder,  surmounted  by  a  right 
circular  cone,  if  the  diameter  of  the  tent  is  100  ft.,  the  height  of  the 
vertical  wall  15  ft.,  and  the  height  of  the  highest  point  of  the  tent  50  ft.  ? 

Ex.  30.  The  diameter  of  the  base  of  a  right  circular  cone  is  equal  to 
its  altitude.    Determine  its  lateral  and  total  area. 


Proposition  YIII.     Theorem 

615.  The  volume  of  a  circular  cone  is  equal  to  the 
area  of  its  base  multiplied  hy  one  third  the  length  of 
its  altitude. 

Hypothesis.     F'=  the  volume,  ^  =  the  area  of  the  base,  and 
H  =  the  altitude  of  a  circular  cone. 
Conclusion.  V=  J  BH. 

Proof  left  to  the  pupil. 
Suggestion.  — Model  the  proof  after  that  in  §  601 .    Use  Fig.  §  613 

616.  Cor.  //  V  denotes  the  volume,  H  the  altitude,  and  R 
the  radius  of  the  base  of  a  circular  cone, 

F=  i  ttR'H. 

Ex.  31.  Determine  the  volume  of  a  right  circular  cone,  the  radius  of 
whose  base  is  7  in.  and  whose  slant  height  is  25  in. 

Ex.  32.  Determine  the  volume  of  the  solid  generated  when  a  right 
triangle  of  base  b  and  altitude  h 

(a)  revolves  about  its  side  h;  (h)  revolves  about  its  side  b. 

Ex.  33.  Determine  the  ratio  of  a  circular  cone  and  a  circular  cylin- 
der having  the  same  base  and  altitude. 


MEASURING  THE   CONE  403 

Proposition  IX.     Theorem 

617.  The  lateral  area  of  a  frustum  of  a  right  circu- 
lar  cone  is  equal  to  the  sum  of  the  circumferences  of  its 
bases,  multiplied  by  one  half  its  slant  height. 


Hypothesis.  S  =  the  lateral  area,  C  and  c  the  circumfer- 
ences of  the  lower  and  upper  bases  respectively,  and  L  =  the 
slant  height  of  a  frustum  of  a  right  circular  cone. 

Conclusion.  S  =  ^L(0-\-c). 

Proof.  1.  Let  aS"  =  the  lateral  area,  C  and  c'  the  circum- 
ferences of  the  lower  and  upper  bases,  and  L'  =  the  slant 
height  of  the  frustum  of  a  regular  pyramid  inscribed  in  the 
frustum  of  the  cone. 

2.  .'.S'  =  iL'(C+c'),  Why? 

3..  Let  the  number  of  faces  of  the  frustum  of  the  pyramid 
be  increased  indefinitely,  keeping  the  pyramid  always  a  regular 
pyramid.     Then 

S'  =  S,  C'=0,  c'=c,  L'=L.  Why? 

4.  .-.  J  L\C'  H-  cO  =  4  i>(0  -h  c).  §  543,  c 

5.  .'.S=\L{C  +  c).  §543,6 

618.  Cor.  1.  If  S  denotes  the  lateral  area^  L  the  slant  height^ 
and  R  and  r  the  radii  of  the  bases  of  a  frustum  of  a  right  cir- 
cular cone,  /S  =  (2  7ri2  -f  2  irr)  X  4  -L  =  Tr(R  +  r)L. 

619.  Cor.  2.  Jlie  lateral  area  of  a  frustum  of  a  right  drcidar 
cone  is  equal  to  the  circumference  of  a  section  midway  between  the 
bases,  multiplied  by  the  slant  height. 


404  SOLID   GEOMETRY  — BOOK  VIII 

Proposition  X.     Theorem 

620.  The  volume  of  a  frustum  of  a  circular  cone  is 
equal  to  the  stem  of  its  bases  and  the  mean  proportional 
between  them,  multiplied  hy  one  third  the  length  of  its 
altitude. 


Hypothesis.  V=  the  volume,  B  and  h  =  the  areas  of  the 
lower  and  upper  bases  respectively,  and  II  =  the  length  of 
the  altitude  of  a  frustum  of  a  circular  cone. 

Conclusion.  V=  -J  H(B  +  6  +  V^). 

Proof.  1.  Inscribe  in  the  frustum  of  the  cone  a  frustum  of 
a  pyramid  having  its  base  inclosed  by  a  regular  polygon.  Let 
V  =  the  volume,  and  B'  and  b'  =  the  areas  of  the  lower  and 
upper  bases  respectively  of  the  frustum  of  the  pyramid. 

2.  Then  F'  =i  H(B'  +  6'  +  VW).  §  572 

Complete  the  proof  as  in  §  601  and  §  615. 

621.  Cor.  If  V  denotes  the  volume,  H  the  altitude,  and  R 
and  r  the  radii  of  the  bases  of  a  frustum  of  a  circular  cone,  then 

V=\iT{B^+r'+Rr)H. 

Ex.  34.  Find  the  lateral  area,  the  total  area,  and  the  volume  of  a 
frustum  of  a  cone  of  revolution,  the  diameters  of  whose  bases  are  16  in. 
and  6  in.,  and  whose  altitude  is  12  in. 

Ex.  35.  Determine  the  contents  in  quarts  of  a  water  pail  having  the 
form  of  a  frustum  of  a  cone  of  revolution,  if  the  diameters  of  the  bottom 
and  top  are  9  in.  and  12  in.  respectively,  and  the  height  of  the  pail  is  14 
in.     (One  quart  occupies  about  231  cu.  in.) 

Note.  —  Supplementary  Exercises  59-71,  p.  459,  can  be  studied  now. 


TANGENT  PLANES  405 

SUPPLEMENTARY  TOPICS 

A.  Planes  Tangent  to  a  Cylinder  or  a  Cone 

622.  A  plane  is  tangent  to  a  circular  cylinder  or  to  a  circular 
cone  when  it  contains  one  and  only  one  element  of  the  cylinder 
or  of  the  cone. 

Proposition  XI.     Theorem 

623.  A  plane  drawn  throtigh  an  element  of  a  circular  cylinder 
and  a  tangent  to  the  base  at  its  extremity  is  tangent  to  the  cylinder. 


Hypothesis.  AA'  is  an  element  of  the  lateral  surface  of  cir- 
cular cylinder  AB',  line  CD  is  tangent  to  the  base  AB  at  Aj 
and  plane  CD'  is  drawn  through  AA'  and  CD. 

Conclnsion.     CD'  is  tangent  to  the  cylinder. 

Proof.  1.  Let  E  be  any  point  in  plane  CD',  not  in  AA',  and 
draw  through  E  a  plane  II  to  the  bases,  intersecting  CD'  in  line 
EF  and  the  cylinder  in  FII. 

2.  Draw  axis  00' ;  then  00'  is  II  AA'.      Ex.  8,  p.  392. 

3.  Let  the  plane  of  00'  and  A  A'  intersect  the  planes  of  AB 
and  FH  in  radii  OA  and  OF,  respectively. 

4.  Then,  GF II  OA  and  FE  II  AD.  Why  ? 
6.  .-.  Z  GFE  =  Z  OAD.  Why  ? 
6.            .'.  FE±  OF,  and  tangent  to  O  FH.             Prove  it. 


406  SOLID   GEOMETRY  — BOOK  VIII 

7.  Whence,  point  E  lies  outside  the  cylinder. 

8.  .-.  all  points  of  CD',  not  in  AA',  lie  outside  the  cylinder, 
and  CD'  is  tangent  to  the  cylinder. 

Proposition  XII.     Theorem 

624.  A  plane  determined  by  an  elemerit  of  the  lateral  surface 
of  a  circular  cone  and  a  tangeyit  to  the  base  at  its  extremity,  is 
tangent  to  the  cone. 


0 


Hypothesis.  OA  is  an  element  of  the  lateral  surface  of  cir- 
cular cone  OAB,  line  CD  is  tangent  to  base  AB  at  A,  and 
plane  OCD  is  drawn  through  OA  and  CD. 

Conclusion.  OCD  is  tangent  to  the  cone. 

(Prove  that  E  lies  outside  the  cone.) 

Suggestion.  —  Model  the  proof  after  that  of  §  623. 

B.     Similar  Cylinders  and  Cones  of 
Revolution 

625.  Similar  cylinders  of  revolution  are  right  circular  cylin- 
ders generated  by  the  revolution  of  similar  rectangles  about 
homologous  sides  as  axes. 

Similar  cones  of  revolution  are  right  circular  cones  generated 
by  the  revolution  of  similar  right  triangles  about  homologous 
sides  as  axes. 


SIMILAR  CYLINDERS  AND   CONES 


407 


Proposition  XIII.     Theorem 

626.  The  lateral  or  total  areas  of  two  siinilar  cylinders  of 
revolution  are  to  each  other  as  the  squares  of  their  altitudes,  or  as 
the  squares  of  the  radii  of  their  bases ;  and  their  volumes  are  to 
each  other  as  the  cubes  of  their  altitudes,  or  as  the  cubes  of  the 
radii  of  their  bases. 


\H 


.""» 


Hypothesis.  S  and  s  are  the  lateral  areas,  T  and  t  are  the 
total  areas,  V  and  v  are  the  volumes,  //  and  h  are  the  altitudes, 
and  M  and  r  the  radii  of  the  bases,  of  two  similar  cylinders 
of  revolution. 

S      T     H'     R'    _,,   V     IP^^ 

h^       r^* 


Conclusion. 


Proof.     1. 

2. 

3. 


-  =  -—  =  -—,  and  — 
t       h^       r^  V 


Since  the  generating  rectangles  are  similar. 

Why? 


II 

h 


r 


h  '' 


II  +  R 

h  +  r' 


§  255,  §  253 


2z^  =  5x^  =  ^  =  ^'.  Whv^ 

2  7rrh        r       r       7^       h^  ^  ' 


6. 


T_2  7rR(H-\-R)^R^^R_R_II^ 

t         2  irr{h  -f-  r)        r       r       r^       h^ ' 


V     ttRH^R^R^R^IP 


TtrVi 


h^ 


Why? 
Why? 


408  PLANE   GEOMETRY  —  BOOK  VIII 

Proposition  XIV.     Theorem 

627.  The  lateral  or  total  areas  of  two  similar  cones  of  revolu- 
tion are  to  each  other  as  the  squares  of  their  slant  heights,  or  as 
the  squares  of  their  altitudes,  or  as  the  squares  of  the  radii  of 
their  bases;  ayid  their  volumes  are  to  each  other  as  the  cubes  of 
their  slant  heights,  or  as  the  cubes  of  their  altitudes,  or  as  the  cubes 
of  the  radii  of  their  bases. 


Hypothesis.  S  and  s  are  the  lateral  areas,  T  and  t  the  total 
areas,  "Fand  v  are  the  volumes,  L  and  I  are  the  slant  heights, 
H  and  h  are  the  altitudes,  and  R  and  r  the  radii  of  the  bases, 
of  two  similar  cones  of  revolution.     (§  625.) 

Conclusion.     ^=2'=^  =  ^  =  «^,a„di:  =  ^  =  ^=^. 

s       t       P       h^       r^  V       P        h'       7^ 

The  proof  is  left  to  the  pupil ;  model  it  after  that  of  §  626. 

Ex.  36.  At  what  distance  from  the  vertex  of  a  right  circular  cone 
with  altitude  H  must  a  plane  parallel  to  the  base  be  passed  so  that  the 
lateral  area  will  be  bisected  ? 


BOOK   IX 

THE   SPHERE 

628.  A  Spherical  Surface  is  a  closed  surface  all  points  of 
which  are  equidistant  from  a  point  within,  called  the  Center. 

629.  A  Sphere  is  the  solid  bounded  by  a  spherical  surface. 

630.  A  Radius  of  a  sphere,  or  of  its  surface,  is  the  straight 
line  drawn  from  its  center  to  any  point  of  its  surface. 

A  Diameter  of  a  sphere,  or  of  its  surface,  is  the  straight  line 
drawn  through  the  center  having  its  extremities  in  the  surface. 

Proposition  I.     Theorem 

631.  The  intersection  of  a  spherical  surface  and  a 
plane  is  a  circle. 


~~^\ 

0'"'.:!\n 

/^l^."-"'' 

"''•^.S^^^ 

'^^«s 

&'■' 

Hypothesis.     ABC  is  the  intersection  of  the  spherical  surface 
whose  center  is  0  and  a  plane. 

Conclusion.  Curve  ABC  is  a  O 

Proof.     1.  Draw  00'  ±  plane  of  ABC. 

2.   Let  A  and  B  be  any  two  points  in  curve  ABC.     Draw 
OA,  OB,  O'A,  and  O'B. 

Suggestion.  —  Now  prove  that  O'A  =  O'B,  and  ABC  is  a  O. 

409 


410  SOLID   OPTOMETRY  —  BOOK  IX 

632.  A  Great  Circle  of  a  sphere  is  the  in- 
tersection of  its  surface  and  a  plane  passing         /^ 
through  its  center ;  as  O  ABQ.  A,-— - 

A  Small  Circle  of  a  sphere  is  the  intersec-  J-t;:;;;;^ 
tion  of  its  surface  and  a  plane  which  does       \  ^ 
not  pass  through  its  center.  ^^J_-^-^ 

The  Axis  of  a  circle  of  a  sphere  is  the 
diameter  of  the  sphere  which  is  perpendicular  to  the  plane  of 
the  circle  ;  as  axis  POP\ 

The  Poles  of  a  Circle  of  a  sphere  are  the  extremities  of  the 
axis  of  the  circle. 

633.  Cor.  1.  Tlie  axis  of  a  circle  of  a  sphere  passes  through 
the  center  of  the  circle. 

634.  Cor.  2.     All  great  circles  of  a  sphere  are  equal. 

635.  Cor.  3.  Every  great  circle  bisects  the  sphere  and  its 
surface. 

For  if  the  portions  of  the  sphere  formed  by  the  plane  of  the  great  circle 
be  separated,  and  placed  so  that  their  plane  surfaces  coincide,  the  spheri- 
cal surfaces  falling  on  the  same  side  of  this  plane,  the  two  spherical  sur- 
faces will  coincide  throughout  ;  for  all  points  of  either  surface  are  equally 
distant  from  the  center. 

636.  Cor.  4.     Any  tivo  great  circles  bisect  each  other. 

For  the  intersection  of  their  planes  is  a  diameter  of  the  sphere,  and 
therefore  a  diameter  of  each  circle. 

637.  Cor.  5.  Between  any  tivo  points  on  the  surface  of  a 
sphere,  not  the  extremities  of  a  diameter,  one  and  only  07ie  arc  of 
a  great  circle,  less  than  a  semicircle,  can  be  draion. 

For  the  two  points,  with  the  center  of  the  sphere,  determine  a  plane 
which  intersects  the  surface  of  the  sphere  in  the  required  arc. 

638.  Two  spheres  are  equal  ivhen  their  radii  are  equal. 

All  radii  and  diameters  of  the  same  sphere  or  equal  spheres 
are  equal. 

639.  A  spherical  surface  may  be  generated  by  the  revolu- 
tion of  a  semicircle  about  its  diameter  as  an  axis. 

For  all  points  of  such  a  surface  are  equally  distant  from  the  center  of 
the  circle. 


THE   SPHERE  411 

640.  Through  two  points  of  a  spherical  surface,  an  infinity 
of  small  circles  of  the  sphere  can  be  drawn.     Why  ? 

Through  three  points  of  a  spherical  surface,  one  and  only 
one  small  circle  of  the  sphere  can  be  drawn.     Why  ? 

Ex.  1.  Prove  that  a  great  circle  of  a  sphere  which  passes  through  one 
pole  of  a  circle  must  pass  through  the  other  pole  also. 

Ex.  2.  How  many  great  circles  can  be  passed  through  two  points 
which  are  the  extremities  of  a  diameter  of  a  sphere  ? 

Ex.  3.  What  kind  of  circles  of  the  earth  are  the  parallels'  of  latitude  ? 

Ex.  4.   What  kind  of  circles  of  the  earth  are  the  meridians  ? 

Ex.  5.   What  kind  of  circle  of  the  earth  is  the  equator  ? 

Ex.  6.  Speaking  strictly,  is  it  accurate  to  speak  of  the  North  Pole  of 
the  earth  ?    Of  what  circle  or  circles  is  it  a  pole  ? 

Ex.  7.  Into  how  many  parts  do  two  great  circles  of  a  sphere  divide 
the  surface  of  the  sphere  ? 

Ex.  8.  Into  how  many  parts  do  three  great  circles  of  a  sphere  divide 
the  surface  of  the  sphere,  if  they  do  not  all  have  a  common  diameter  ? 

Ex.  9.  Prove  that  all  circles  of  a  sphere  made  by  parallel  planes  have 
the  same  axis  and  the  same  poles. 

Ex.  10.  Given  a  point  of  a  spherical  surface.  Prove  that  it  is  the 
pole  of  one  and  only  one  great  circle. 

Ex.  11.  Prove  that  circles  of  a  sphere  made  by  planes  equidistant 
from  the  center  of  the  sphere  are  equal. 

Suggestion.  —  Use  the  Pythagorean  Theorem.    (§  291.) 

Ex.  12.    State  and  prove  the  converse  of  Ex.  11. 

Ex.  13.  Prove  that  circles  of  a  sphere  made  by  planes  unequally  dis- 
tant from  the  center  of  the  sphere  are  unequal,  the  more  remote  being 
the  smaller. 

Ex.  14.   State  and  prove  the  converse  of  Ex.  13. 

Ex.  15.  In  how  many  points  can  two  straight  lines  intersect  ? 

In  how  many  points  on  one  hemisphere  can  two  great  circles  intersect  ? 

641.  The  great  circles  of  a  sphere  in  sphencal  geometry  cor- 
respond to  the  straight  lines  of  a  plane  in  plane  geometry. 
§  637  and  Ex.  15  are  two  instance's  pointing  to  this  similarity 
of  great  circles  and  straight  lines ;  others  will  appear  in  the 
remaining  paragraphs  of  Book  IX. 


412  SOLID  GEOMETRY  —  BOOK  IX 

642.  It  can  be  proved  that  the  length  of  the  arc  of  the  great 
circle,  less  than  a  semicircle,  between  two  points  of  a  spherical 
surface  is  less  than  the  length  of  any  other  curved  line  on  the 
surface  between  the  two  points.  Consequently,  the  distance 
between  two  points  on  the  surface  of  a  sphere^  measured  on  the 
surface,  is  defined  to  be  the  arc  of  the  great  circle,  less  than  a 
semicircle,  drawn  between  them. 

Proposition  II.     Theorem 

643.  All  points  in  a  circle  of  a  sphere  are  equidistant 
from  each  of  its  poles. 


•p' 

Hypothesis.     P  and  P'  are  the  poles  of  O  ABC  of  sphere  M. 

Conclusion.  All  points  of  O  ABC  are  equidistant  from  P, 
and  also  from  P'. 

Proof.  1.  Let  A  and  B  be  any  two  points  of  O  ABC,  and 
draw  great  circle  arcs  PA  and  PB.  Draw  axis  PMP',  intersect- 
ing the  plane  of  ABC  at  0.     Draw  OA,  OB,  PA,  and  PB. 

2.  .'.PA  =  PB.  Prove  it. 

3.  .♦.  PA  =  PB.  Prove  it. 

4.  Since  A  and  B  are  any  two  points  of  O  ABC,  .'.  all 
points  of  O  ABC  are  equidistant  from  P. 

5.  Similarly  all  points  of  O  ABC  are  equidistant  from  P\ 

644.  The  Polar  Distance  of  a  circle  of  a  sphere  is  the  dis- 
tance from  the  nearer  of  its  poles  to  the  circle,  or  from  either 
pole  if  they  are  equally  near. 

Thus,  in  the  figure  of  Proposition  II,  the  polar  distance  of  O  ABC  is 
arc  PA. 


THE   SPHERE 


413 


645.  Cor.  All  points  of  a  great  circle  of  a  sphere  are  at  a 
quadranfs  distance  from  either  of  its  poles. 

Note.  —  The  term  quadrant  in  Spherical  Geom- 
etry, usually  signifies  a  quadrant  of  a  great  circle. 

Hypothesis.  P  is  a  pole  of  great  circle  ABC  of 
sphere  AFC;  B  is  any  point  in  QABC,  and  PB  is 
an  arc  of  a  great  O. 

Conclusion.     Arc  PB  =  a  quadrant. 

Swjgestion. — Draw  radii  OA,  OB,  and  OP. 

Note.  —  An  arc  of  a  circle  may  be  drawn  on  the  surface  of  a  sphere  by 
placing  one  foot  of  the  compasses  at  the  nearer  pole  of  the  circle,  the  dis- 
tance between  the  feet  being  equal  to  the  chord  of  the  polar  distance. 


Proposition  III.     Theorem 

646.  A  point  on  the  surface  of  a  sphere  at  a  quad- 
rant's distance  from  each  of  two  points^  not  the  extremi- 
ties of  a  diameter  of  the  sphere,  is  a  pole  of  the  great 
circle  through  those  points. 


Hypothesis.  P  is  on  the  surface  of  the  sphere  whose  center 
is  0.  AB  is  an  arc  of  great  O  ABC,  not  a  semicircle.  PA 
and  PB  are  quadmnts. 

Conclusion.  P  is  a  pole  of  AB. 

Suggestions.  —  1.  Recall  the  definition  of  "  pole  of  a  0." 
2.  Draw  PO,  AO,  OB,  and  prove  PO  1  plane  ABC. 

"Ex.  16.  If  a  point  lies  at  a  quadrant^s  distance  from  the  ends  of  a 
diameter  of  a  sphere,  is  it  necessarily  a  pole  of  the  great  circle  through 
those  points  ? 


414 


SOLID   GEOMETRY  —  BOOK   IX 


647.  The  angle  between  two  intersecting  curves  is  the  angle 
formed  by  the  tangents  to  the  curves  at  the  point  of  inter- 
section. 

A  Spherical  Angle  is  the  angle  between  two  intersecting  arcs 
of  great  circles. 


Propositiois^  IV.     Theorem 

648.  A  spherical  angle  is  measured  by  an  arc  of  a 
great  circle  having  its  vertex  as  a  pole,  included  be- 
tween its  sides  extended  if  necessary. 


Hypothesis.  ABO  and  AB'C  are  arcs  of  great  (D  on  the 
surface  of  the  sphere  whose  center  is  0 ;  lines  AD  and  AD' 
are  tangent  to  ABC  and  AB'C,  respectively,  and  BB'  is  an  arc 
of  a  great  O  having  ^  as  a  pole,  included  between  arcs  ABC 
and  AB'C. 

Conclusion.     Z  BAB'  is  measured  by  arc  BB'. 

Proof.     1.   Draw  diameter  AGO  and  radii  OB  and  OB'. 

2.  Arcs  AB  and  AB'  are  quadrants.  Why  ? 

3.  .-.  A  AOB  and  AOB'  are  rt.  A.  Why  ? 

4.  OB  II  AD  and  OB'  II  AD'.  Why? 

5.  .-.  Z  DAD'  =  Z  BOB'.  §  481 

6.  But  Z  BOB'  is  measured  by  arc  BB'.  Why  ? 

7.  Then,  Z  DAD'  is  measured  by  arc  BB'.  Why  ? 

8.  .-.  Z  £^5'  is  measured  by  arc  BB'.  §  647 


THE   SPHERE  415 

649.  Cor.  1.  Tlie  angle  between  two  arcs  of  great  circles  is 
equal  to  the  diedral  angle  formed  by  their  planes. 

650.  Cor.  2.  An  arc  of  a  great  circle  drawn  to  another  great 
circle  from  the  latter^ s  pole  is  popendicular  to  that  great  circle. 

Suggestions.— 1.  What  Z  does  OA  form  with  plane  BOB'? 
2.  What  kind  of  Z  is  diedral  Z  AOB'B?    (§  495.) 

Ex.  17.  If  a  spherical  blackboard  can  be  had,  construct  a  spherical 
angle  and  measure  it. 

Ex.  18.  The  chord  of  the  polar  distance  of  a  circle  of  a  sphere  is  6. 
If  the  radius  of  the  sphere  is  6,  what  is  the  radius  of  the  circle  ? 

Ex.  19.  What  is  the  locus  of  points  in  space  at  a  given  distance  d  from 
a  fixed  point,  and  equidistant  from  two  given  points  ? 

SPHERICAL   POLYGONS 

Note.  —  Recall  at  this  point  the  definition  of  a  polygon  in  plane  geom- 
etry as  a  closed  broken  line  lying  in  a  plane.     (§  125.) 

Recall  also  §  641,  calling  attention  to  the  similarity  in  the  rdles  of  the 
straight  line  in  plane  geometry  and  the  great  circle  in  solid  geometry. 

651.  A  Spherical  Polygon  is  a  closed  line  on  the  surface  of  a 
sphere  consisting  of  arcs  of  three  or  more 
great  circles  ;  as  polygon  A  BCD. 

Just  as  in  plane  geometry  we  considered 
only  convex  polygons,  so  we  shall  consider 
only  convex  spherical  polygons.    (See  §  126.) 

It  will   be  assumed   as   evident   that  a 
simple   spherical   polygon   iucloses    a    por- 
tion of  the   surface   of  the   sphere.      The 
bounding  arcs  are  the  Sides  of  the  polygon ;  they  are  usually 
measured  in  arc-degrees.     (§  214.) 

The  angle  formed  by  two  consecutive  sides  of  a  polygon  is 
an  Angle  of  the  spherical  polygon,  and  its  vertex  is  a  Vertex  of 
the  polygon. 

A  Diagonal  of  a  spherical  polygon  is  the  arc  of  the  great 
circle  joining  two  non-consecutive  vertices  of  the  polygon,  and 
lying  withia  the  polygon. 


416  SOLID   GEOMETRY  —  BOOK   IX 

652.  A  Spherical  Triangle  is  a  spherical  polygon  having 
three  sides.  A  spherical  triangle  is  Isosceles  when  it  has  two 
equal  sides  ;  it  is  Equilateral  when  it  has  three  equal  sides ;  it 
is  Right-angled  when  one  of  its  angles  is  a  right  angle. 

653.  The  planes  of  the  sides  of  a  spherical  polygon  form  a 
polyedral  angle,  whose  vertex  is  the  center  of  the  sphere,  and 
whose  face  angles  are  measured  by  the  sides  of  the  spherical 
polygon. 

Thus,  in  the  figure  of  §  651,  the  planes  of  the  sides  of  the  spherical 
polygon  form  a  polyedral,  Z  0-ABCD,  whose  face  angles  AOB,  BOC, 
etc.,  are  measured  by  arcs  AB,  BC,  etc.,  respectively. 

Ex.  20.  Prove  that  the  angles  of  a  spherical  polygon  have  the  same 
measures  as  the  diedral  angles  of  the  corresponding  polyedral  angle. 

654.  If  great  circles  be  drawn  with  the  vertices  of  a  spheri- 
cal triangle  as  poles,  they  divide  the  surface  of  the  sphere  into 
eight  parts  whose  boundaries  are  tri- 
angles. 

Thus,  if  circle  B'C'B"  be  drawn  with 
vertex  A  of  spherical  A  ABC  as  a  pole, 
circle  A'C^A"  with  ^  as  a  pole,  and  circle 
A'Bf'A"B'  with  (7  as  a  pole,  the  surface 
of  the  sphere  is  divided  into  eight  spheri- 
cal A  ;  namely,  A'B'C,  A'B"Q',  A'B'C, 
and  A'B"C'  on  the  hemisphere  represented  in  the  figure,  and 
four  others  on  the  opposite  hemisphere. 

Of  these  eight  spherical  A,  one  is  called  the  Polar  Triangle 
of  ABO,  and  is  determined  as  follows  : 

Of  the  intersections.  A!  and  A'  of  circles  drawn  with  B  and 
C  as  poles,  that  which  is  nearer  to  A,  i.e.  A',  is  a  vertex  of  the 
polar  triangle ;  and  similarly  for  the  other  intersections. 

Thus,  A'B'C  is  the  polar  A  of  ABC. 

Ex.  21.  The  polar  distance  of  a  circle  of  a  sphere  is  60°.  If  the  di- 
ameter of  the  circle  is  6,  find  the  diameter  of  the  sphere,  and  the  distance 
of  the  circle  from  its  center. 

Suggestion.  —  Represent  the  radius  of  the  sphere  by  2  x.     (§  288.) 


SPHERICAL  POLYGONS  417 

Proposition  V.     Theorem 

655.  If  one  spherical  triangle  is  the  polar  triangle 
of  another,  then  the  second  sjjherical  triangle  is  the 
polar  triangle  of  the  first. 


Hypothesis.  A'B'C  is  the  polar  A  of  the  spherical  A  ABC; 
Af  B,  and  C  are  the  poles  of  arcs  B'C\  O'A',  and  A'B', 
respectively. 

Conclusion.     ABC  is  the  polar  A  of  spherical  A  A'BfC. 

Proof.     1.   A'  is  at  a  quadrant's  distance  from  B.  §  645 
(Since  B  is  the  pole  of  arc  A'C.) 

2.  A'  is  at  a  quadrant's  distance  from  C.  Why  ? 

3.  .-.  A'  is  a  pole  of  the  great  O  arc  BC.  Why  ? 

4.  Similarly  Bf  and  C  are  the  poles  oi  AC  and  AB, 
respectively.  Prove  it. 

5.  .-.  A  ABC  is  the  polar  A  of  A  A'BfC. 

(For  of  the  two  intersections  of  the  great  (D  having  B^  and 
C,  respectively,  as  poles,  A  is  nearer  to  A' ;  similarly  for  B 
and  C.  §  654 

Note.  — Two  spherical  triangles,  each  of  which  is  the  polar  triangle  of 
the  other,  are  called  polar  triangles. 

Ex.  22.  How  many  degrees  are  there  in  the  polar  distance  of  a  circle 
whose  plane  is  6V2  units  from  the  center  of  the  sphere,  the  diameter  of 
the  sphere  being  20  units  ? 

Suggestion.  —  The  radius  of  the  0  is  a  leg  of  a  rt.  A,  whose  hypotenuse  is 
the  radius  of  the  sphere,  and  whose  other  leg  is  the  distance  from  its  center 
to  the  plane  of  the  0. 


418  SOLID   GEOMETRY  — BOOK  IX 

Proposition  VI.     Theorem 

656.  In  two  polar  triangles,  each  angle  of  one  has  the 
same  measure  as  the  supplement  of  that  side  of  the  other 
of  which  it  is  the  pole. 


Hypothesis.     A  ABC  and  A'B'C  are  polar  A,  point  A  being 

the  pole  of  B^',  etc. 

Let  a,  a',  etc.  be  the  measures  in  degrees  of  BO,  B'C,  etc., 
respectively ;  let  A,  A',  etc.  be  the  measures  in  degrees  of 
A  A,  A',  etc. 

Conclusion.      ^  =  180  -  a' ;  JB  =  180  -  &' ;  (7  =  180  -  c'. 
^'  =  180  -  a ;  B'==  180  -b  ;    C  =  180  -  c. 

Proof.  1.  Extend  AB  and  AC  to  meet  B^  at  D  and  E, 
respectively. 

2.  Since  B'  is  the  pole  of  AE,  B'E  =  90°.  Why  ? 

3.  Similarly,  CD  =  90°. 

4.  .-.  B'E  +  CI)  =  1S0°. 

5.  .'.  m)  +  DE-^CD  =  180°,  OY  DE-{-B^' =  1S0°. 

6.  But  DE  is  the  measure  of  Z  ^.  §  648 

7.  .-.  A-\-a'  =  180,  or 

^=180 -a'. 

8.  Similarly  for  each  of  the  other  angles  of  either  triangle. 

Ex.  23.     Prove  on  the  figure  for  §  656,  that  A'  =  180  —  a. 

Ex.  24.  If  the  sides  of  a  spherical  triangle  are  77°,  123^,  and  95°, 
how  many  degrees  are  there  in  each  angle  of  its  polar  triangle  ? 

Ex.  25.  If  the  angles  of  a  spherical  triangle  are  86°,  131°,  and  68°, 
how  many  degrees  are  there  in  each  side  of  its  polar  triangle  ? 


SPHERICAL  POLYGONS 


419 


Proposition  VII.     Theorem 

657.   Tlie  sum  of  the  sides  of  a  convex  spherical  poly- 
(jon  is  less  than  360°. 


Hypothesis.     ABCD  is  a  convex  spherical  polygon. 

Conclusion.     AB  +  BC+ CD -\- DA  <  360°. 

Proof.  1.  Let  polyedral  angle  0-ABCD  be  the  polyedral 
angle  which  corresponds  to  spherical  polygon  ABCD. 

2.  Then  the  measure  of  AB  equals  the  measure  of  central 
angle  AOB.  Why  ? 

3.  Similarly  for  arcs  BC,  CD,  and  DA. 

4.  But  Z  AOB  +  Z  BOC  +  Z  COD  +  Z  DO  A  <  360°. 

_       ^       ^        ^  §614 

5.  .-.  AB  +  BC-^CD-\-DA<  360°. 

Note  I.  —  The  pupil  should  recall  at  this  point  that  one  arc-degree  is  ^^^ 
of  a  circle.  Since  arcs  AB,  BC,  CD,  and  DA  are  arcs  of  great  circles, 
Proposition  VII  means  that  the  sum  of  the  sides  of  any  spherical  polygon 
is  less  than  300  arc-degrees  of  a  great  circle  —  i.  e.  is  less  than  a  great 
circle. 

Note  2.  —  Proposition  VII  is  one  of  many  theorems  about  spherical 
polygons  which  can  be  formulated  from  corresponding  theorems  about 
polyedral  angles  by  replacing  in  the  latter  the  words  "face  angle"  and 
"diedral  angle"  by  "side"  and  "angle"  respectively. 

Ex.  26.  If  two  sides  of  a  spherical  triangle  measure  80°  and  70°  respec- 
tively, between  what  two  values  must  the  remaining  side  lie  ? 

Ex  27.  What  is  the  greatest  possible  length  for  the  sum  of  the  sides 
of  a  convex  spherical  polygon  on  a  circle  of  radius  12  inches  ? 


420  SOLID   GEOMETRY  — BOOK  IX 

Proposition  VIII.     Theorem 

658.   Tlie  sum  of  the  angles  of  a  spherical  triangle  is 
greater  than  180°  and  less  than  540°. 


Hypothesis.  A,  B,  and  C  are  tlie  measures  in  degrees  of 
the  A  of  spherical  A  ABC. 

Conclusion.  A  +  B  +  C>  180°  and  <  540°. 

Proof.     1.     Let  A  A'B'C  be  the  polar  triangle  of  spherical 

A  ABC,  A  being  the  pole  of  B^,  B  of  A^,  and  C  of  A^'. 

Let  the  measures  in  degrees  of  B'C,  C'A',  and  A'B'  be  a" 
b',  and  c',  respectively. 

2.  .-.  ^  =  180  -  a',  5  =  180  -  6',  and   0  =  180  -  c'.     Why  ? 

3.  .♦.  ^  +  ^H-C=540-(a'+6'  +  c').  Why? 

4.  .-.  ^-h^+C<540°. 

5.  But  a'  +  b'  +  c'  <  360°.  §  657 

6.  .'.A  +  B+C>  180°.  Ax.  20,  §  158 

659.  Cor.  Tlie  sum  of  the  angles  of  a  spherical  polygon  of  n 
sides  is  greater  than  (n  —  2)  x  180°. 

Consider  ABCD  a  spherical  quadrilateral. 

Draw  great  O  arc  BD,  dividing  the  quadrilateral  into  two  spherical  i^, 
ABD  and  BBC.  The  sum  of  the  angles  of  the  triangles  equals  the  sum 
of  the  angles  of  the  quadrilateral.  In  each  triangle  the  sum  of  the  angles 
is  >  180°.  Hence  for  the  quadrilateral,  the  sum  of  the  angles  is  greater 
than  2  X  180^ ;  that  is,  the  sum  >(4  -  2)  x  180°. 

In  like  manner,  if  there  are  n  sides,  the  polygon  can  be  divided  into 
(n— 2)  spherical  triangles,  in  each  of  which  the  sum  of  the  angles  is 
gi-eater  than  180°.  Therefore,  in  the  polygon  the  sum  of  the  angles  is 
greater  than  (n  —  2)  x  180°. 


SPHERICAL  POLYGONS  421 

660.  The  Spherical  Excess  of  a  spherical  triangle,  measured 
in  degrees,  is  the  diii'erence  between  the  sum  of  its  angles  and 
180°. 

The  Spherical  Excess  of  a  spherical  polygon  of  n  sides,  meas- 
ured in  degrees,  is  the  difference  between  the  sum  of  its  angles 
and  (n  -  2)  X  180°. 

Note. —  In  each  case,  the  spherical  excess  is  the  amount  by  which  the 
sum  of  the  angles  of  the  spherical  polygon  exceeds  the  sum  of  the  angles 
of  a  plane  polygon  of  the  same  number  of  sides. 

Ez.  28.  Prove  that  a  spherical  triangle  may  have  one,  two,  or  three 
right  angles,  or  one,  two,  or  three  obtuse  angles. 

661.  A  spherical  triangle  having  two  right  angles  is  called 
a  Bi-rectangular  Triangle,  and  one  having  three  right  angles 
a  Tri-rectangular  Triangle. 

Ex.  29.  Prove  that  the  sum  of  the  angles  of  a  spherical  hexagon  is 
greater  than  8,  and  less  than  12,  right  angles. 

Ex.  30.  What  is  the  spherical  excess  of  a  triangle  whose  angles  are 
100°,  95°,  and  65^  respectively  ? 

Ex.  31.     What  is  the  spheiical  excess  of  a  tri-rectangular  triangle  ? 

Ex.  32.  Prove  that  the  spherical  excess  of  a  bi-rectangular  triangle 
is  the  measure  of  the  remaining  angle  of  the  triangle.     (§  048.) 

Ex.  33.  What  is  the  spherical  excess  of  a  triangle  if  the  sides  of  its 
polar  triangle  measure  80°,  85°,  and  95°  ? 

Ex.  34.  What  relation  exists  between  a  tri-rectangular  spherical 
triangle  and  its  polar  ? 

Ex.  35.  Prove  that  the  sides  opposite  the  equal  angles  of  a  bi-rectaii- 
gular  triangle  are  quadrants. 

Suggestion.  —  Recall  §  499  and  the  definition  of  "pole." 

Ex.  36.     Prove  that  each  side  of  a  tri-rectangular  triangle  is  a  quadrant. 

Ex.  37.  Prove  that  in  a  bi-rectangular  spherical  triangle,  the  third 
angle  has  the  same  measure  as  the  side  opposite  it. 

662.  If  a  ^lane  or  a  line  has  only  one  point  in  common 
with  the  surface  of  a  sphere,  it  is  said  to  be  Tangent  to  the 
Sphere.  The  sphere  is  said  to  be  tangent  to  the  plane  or  line. 
The  point  common  to  the  plane  or  line  and  the  spherical  sur- 
face is  called  the  Point  of  Contact  or  Tangency. 


422 


SOLID   GEOMETRY  — BOOK  IX 


Proposition  IX.     Theorem 

663.    A  plane  perpendicular  to  a  radius  of  a  sjjJiere 
at  its  outer  extremity  is  tangent  to  the  sphere. 


^^^^ 

\y 

f 

>         /           y 

y^\ 

,' 

\    y     X 

/  / 

~— L-^^       / 

/     ^ 

A       y 

M 


Hypothesis.     Plane  MN~  ±  radius  OA  at  A. 
Conclusion.     Plane  MN  is  tangent  to  the  sphere. 

Suggestion.  —  Let  B  be  any  poiut  of  MN  except  A. 
Prove  that  B  lies  outside  the  sphere. 

664.  Cor.  A  plane  tangent  to  a  sphere  is  perpendicular  to  the 
radius  drawn  to  the  point  of  contact.     (Fig.  of  Prop.  IX.) 

665.  Two  Spheres  are  Tangent  to  each  other  when  each  is 
tangent  to  the  same  plane  at  the  Same  point. 

Ex.  38.  Prove  that  a  straight  line  perpendicular  to  a  radius  of  a 
sphere  at  its  outer  extremity  is  tangent  to  the  sphere. 

Ex.  39.  How  many  straight  lines  can  be  tangent  to  a  sphere  at  a 
point  of  the  sphere  ? 

Ex.  40.  If  two  straight  lines  are  tangent  to  a  sphere  at  the  same  point, 
their  plane  is  tangent  to  the  sphere. 

Ex.  41.  Prove  that  all  lines  tangent  to  a  sphere  at  a  point  of  the 
sphere  lie  in  the  plane  tangent  to  the  sphere  at  that  point. 

Ex.  42.  How  many  straight  lines  can  be  tangent  to  a  sphere  from  a 
point  outside  the  sphere  ?     Compare  the  lengths  of  these  tangents. 

Ex.  43.  Prove  that  the  points  of  contact  of  all  lines  tangent  to  a 
sphere  from  an  exterior  point  lie  in  a  circle. 

Ex.  44.    Is  the  circle  in  Ex.  43  a  great  circle  or  a  small  circle  ? 

Ex.  45.  If  two  spheres  are  tangent  to  the  same  plane  at  the  same 
point,  the  straight  line  joining  their  centers  passes  through  the  point  of 
contact. 


MEASUREMENT   OF  SPHERICAL  POLYGONS     423 


MEASUREMENT  OF  SPHERICAL  POLYGONS 

666.  A  Zone  is  the  portion  of  a  spherical  surface  included 
between  two  parallel  planes. 

The  circles  which  bound  the  zone  are  its  baseSj  and  the  dis- 
tance between  their  planes  is  its  dUitude. 

A  zone  of  one  base  is  a  zone  lying  between  one  plane  and 
a  parallel  plane  tangent  to  the  sphere. 

667.  If  semicircle  ACEB  be  revolved 
about  diameter  AB  as  an  axis,  and  CD  and 
EF  are  lines  ±  AB,  then  arc  CE  generates 
a  zone  whose  altitude  is  DF,  and  arc  AC  a. 
zone  of  one  base  whose  altitude  is  AD. 

668.  Application  of  Limits  to  Zones. 

Let  0  be  the  center  of  AGB,  and  OMhe  any  diameter  of  the 
circle.  Let  AA'  and  BB'  be  _L  OM.  Let  C  bisect  arc  AB,  and 
draw  broken  line  ACB.     If  arc  AB  is  ,, 

revolved  about  OM  as  axis,  it  gener- 
ates a  zone,  and  broken  lines  AC  and 
CB  generate  curved  surfaces. 

It  will  be  assumed  as  evident  that 
the  sum  of  the  areas  of  the  surfaces 
generated  by  AC  and  CB  is  less  than 

the  area  of  the  zone  generated  by  ACB. 

Assume  arcs  AC  and  CB  to  be  bisected  at  X  and  Y",  and 
imagine  the  broken  line  AXCYB.  It  will  be  assumed  as  evi- 
dent that  the  area  of  the  surface  generated  by  line  AXCYB  is 
greater  than  that  generated  by  ACB  but  is  still  less  than  that 
of  the  zone  generated  by  AB.  If  the  process  of  subdividing 
arc  AB  by  successively  halving  the  subdivisions  of  arc  AB  be 
continued  indefinitely,  it  will  be  assumed  evident  that  the  sur- 
face generated  by  the  resulting  broken  line  approaches  as  limit 
the  zone  AB  ;  also,  as  the  chords  like  AC  decrease  indefinitely 
in  length,  their  distance  from  center  0  increases  and  approaches 
the  radius  of  arc  AB  as  limit. 


424 


SOLID   GEOMETRY  — BOOK   IX 


Proposition  X.     Theorem 

669.  The  area  of  the  surface  generated  hy  the  revolu- 
tion of  a  straight  line  about  a  straight  line  in  its  plane, 
not  parallel  to  a7id  not  intersecting  it,  as  an  axis,  is 
equal  to  its  projection  on  the  axis,  multiplied  hy  the  cir- 
cumference of  a  circle,  whose  radius  is  the  perpendicu- 
lar erected  at  the  mid-point  of  the  line  and  terminating 
in  the  axis. 


Hypothesis.  Straight  line  AB  is  revolved  about  straight  line 
FM  in  its  plane,  not  ±  to  and  not  intersecting  it,  as  an  axis ; 
lines  AC  and  BD  ±  FM,  and  EF  is  the  ±  erected  at  the  mid- 
point of  AB  terminating  in  FM. 

Conclusion.         Area  AB'^=CDy.2 irEF. 

Proof.     1.   Draw  line  AG  ±  BD,  and  line  EH  ±  CD. 

2.  The  surface  generated  by  AB  is  the  lateral  surface  of  a 
frustum  of  a  cone  of  revolution,  whose  bases  are  generated  by 
AC  and  BD. 

3.  .-.  area  AB=ABx2  ttEH 

4.  A  ABG  and  EFH  are  similar. 
g  . AB^EF 

'  '  AG     EH 
6.  .\ABxEH  =  AG  xEF 

=  CDx  EF. 
1.   Substituting  in  Step  3, 

area  AB=CDx2  ttEF. 


§619 
Prove  it. 

Why? 

Why? 
Why? 


*The  expression 
generated  by  AB. 


area  AB  "  is  used  to  denote  the  area  of  the  surface 


MEASUREMENT  OF  SPHERICAL  POLYGONS     425 

Proposition  XI.     Theorem 

670.   The  area  of  a  zone  is  equal  to  its  altitude  multi- 
plied by  the  circumference  of  a  great  circle. 


Hypothesis.  AB  is  revolved  about  diameter  OM  as  axis ; 
AA'  and  BB'  ±  OM',  R  is  the  radius  of  AB. 

Conclusion.     Area  of  zone  generated  by  AB  =  ^'jB'  x  2  irR. 

Proof.  1.  Bisect  ^^  at  C;  draw  AG  and  GB]  also  draw 
CC  J.  03f  and  OELAG. 

2.  Revolve  the  ligure  about  OiHf  as  axis. 

3.  .•.area^a  =  ^'C"  x2  7rO^.  §669 

area  GB  =  O'jB'  x  2  ttOE.  Why  ? 

4.  Adding,  the  surface  generated  by  AGB 

=  (A'G'  +  CB')x2  7rOE. 

5.  Continue  to  bisect  the  subdivisions  of  AB,  indefinitely. 

6.  Then,  the  area  of  the  surface  generated  by  revolving  the 
inscribed  broken  line  approaches  as  limit  the  area  of  the  zone 
generated  by  AB,  and 


OE=R. 
'.  area  of  zone  =  A'B'  x2  wR. 


§668 
§403* 


Note.  — The  proof  of  §  670  holds  for  auy  zone  which  lies  entirely  on  the 
surface  of  a  hemisphere ;  for,  in  that  case,  no  chord  is  II  OM,  and  §  669  is 
applicable. 

Since  a  zone  which  does  not  lie  entirely  on  the  surface  of  a  hemisphere 
may  be  considered  as  the  sum  of  two  zones,  each  of  which  does  lie  entirely 
on  the  surface  of  a  hemisphere,  the  theorem  of  §  670  is  true  for  any  zone. 


426 


SOLID   GEOMETRY  — BOOK   IX 


671.  Cor.  1.  If  Z  denotes  the  area  of  a  zone,  h  its  altitude, 
ayid  R  the  radius  of  the  sphere^ 

Z=2  irBh. 

672.  Cor.  2.  The  area  of  a  spherical  surface  equals  the 
square  of  its  radius  multiplied  by  4  tt. 

Proof.  A  spherical  surface  may  be  regarded  as  a  zone  whose  altitude 
is  a  diameter  of  the  sphere.  Letting  S  represent  the  area  of  the  spherical 
surface, 

673.  Cor.  3.  The  area  of  the  surface  of  a  sphere  equals  the 
area  of  four  great  circles  of  the  sphere. 

674.  Cor.  4.  The  areas  of  two  spherical  surfaces  have  the 
same  ratio  as  the  squares  of  their  radii  or  the  squares  of  their 
diameters. 

675.  A  Lune  is  the  portion  of  a  spherical  surface  bounded 
by  two  semicircles  of  great  circles ;  as  ACBD. 

The  Angle  of  a  Lune  is  the  angle  between 
its  bounding  arcs. 

It  is  evident  that  two  lunes  on  the  same 
sphere  or  equal  spheres  are  congruent  if  their 
angles  are  equal. 

676.  It  is  evident  that  lunes  on  the  same 
sphere  or  on  equal  spheres  may  be  added  by 
placing  them  so  that  their  angles  become  adja- 
cent angles  ;  thus  lune  ACBE  +  lune  AEBF 
=  lune  ACBF. 

When  two  lunes  are  added,  the  angle  of  the 
sum  equals  the  sum  of  the  angles  of  the  given 
iunes. 

If  Lj^  is  used  to  denote  the  lune  whose  angle  is  Z  X,  then 

Lj[-{-  Ly=  L(^x+r)  5 

i.e.  lune  of  Z  X  +  lune  of  Z  F=  lune  of  Z(X  +  Y). 

Note.  —  It  is  very  important  that  the  symbol  Lx  he  understood  and 
remembered. 


MEASUREMENT  OF   SPHERICAL  POLYGONS     427 


Proposition  XII.     Theorem 

677.   Two  lunes  on  the  same  sphere  or  equal  spheres 
ham  the  same  ratio  as  their  angles. 

Case  I.     When  the  angles  are  commensurable. 


Hjrpothesis.  ACBD  and  ACBE  are  lunes  on  sphere  AB, 
having  their  A  CAD  and  CAE  commensurable. 

Conclusion.  ACBD^Z^CAD^ 

ACBE     Z  CAE 

Proof.  1.  Let  Z  CAa  be  a  common  measure  of  Z  CAD  and 
CAE,  and  let  it  be  contained  5  times  in  Z  CAD,  and  3  times 
in  Z  CAE. 

2  .  ZC^D^5  ... 

' '  Z  CAE     3*  ^  ^ 

3.   Extending   the  arcs  of  division   of  Z  CAD  to  JB,  lune 

ACBD  will  be  divided  into  5  parts,  and  lune  ACBE  into  3 

parts,  all  of  which  parts  will  be  equal.  Why  ? 

.  ACBD^5 

"ACBE     3* 


4. 


6.   From  (1)  and  (2), 


ACBD ^Z  CAD 
ACBE     Z  CAe' 


(2) 
Why? 


Note. — The  theorem  may  be  proved  in  a  similar  manner  when  the 
given  lunes  are  on  equal  spheres. 

Case  II.      Wheii  the  angles  are  incommensurahle. 

Suggestion.  —  Model  the  proof  after  that  in  §  544. 


428  SOLID  GEOMETRY  —  BOOK   IX 

678.  The  surface  of  a  hemisphere  may  be  regarded  as  a  lune 
of  angle  180°  and  the  surface  of  the  sphere,  a  lune  of  angle  360°. 

679.  Cor.  1.  The  surface  of  a  lune  is  to  the  surface  of  the 
sphere  as  the  measure  of  its  angle  in  degrees  is  to  360. 

680.  Cor  2.  If  the  radius  of  the  sphere  is  R,  and  the  angle  of 
the  lune,  measured  in  degrees,  is  A,  and  the  area  of  the  lune  is  de- 
noted hy  L^,  then 

^      360  90 

Ex.  46.  Prove  that  the  areas  of  two  zones  on  the  same  sphere,  or 
equal  spheres,  are  to  each  other  as  their  altitudes. 

Ex.  47.  Determine  the  area  of  a  zone  whose  altitude  is  13,  if  the 
radius  of  the  sphere  is  16. 

Ex.  48.  Prove  that  the  area  of  a  zone  of  one  base  is  equal  to  the  area 
of  the  circle  whose  radius  is  the  chord  of  its  generating  arc.     (§  288.) 

Ex.  49.  Determine  the  area  of  the  surface  of  a  sphere  whose  radius 
is  12. 

Ex.  50.  If  the  radius  of  a  sphere  is  i?,  what  is  the  area  of  a  zone  of 
one  base,  whose  generating  arc  is  45°  ? 

Ex.  51.  Find  the  radius  of  a  sphere  whose  surface  is  equivalent  to  the 
entire  surface  of  a  cylinder  of  revolution,  whose  altitude  is  lOi,  and  ra- 
dius of  base  3. 

Ex.  52.  What  is  the  area  of  a  lune  whose  angle  is  40°  on  the  surface 
of  a  sphere  whose  radius  is  15  in.  ? 

Ex.  53.  What  part  of  the  surface  of  the  earth  is  included  between 
the  30th  and  35th  meridians  ? 

Ex.  54.  The  area  of  a  lune  is  28f .  If  the  area  of  the  surface  of  the 
sphere  is  120,  what  is  the  angle  of  the  lune  ? 

Ex.  55.  Prove  that  the  surface  of  a  sphere  is  equal  to  two  thirds  the 
entire  surface  of  the  right  circular  cylinder  circumscribed  about  it. 

Ex.  56.  Compare  the  surface  of  a  sphere  with  the  lateral  surface  of 
the  right  circular  cylinder  circumscribed  about  the  sphere. 

Ex.  57,  What  circles  of  the  earth  bound  the  North  Temperate  Zone  ? 
What  part  of  the  earth's  surface  lies  within  that  zone  ? 

Ex.  58.     What  zones  of  the  earth  are  zones  of  one  base  ? 


MEASUREMENT   OF  SPHERICAL  POLYGONS     429 

681.  Two  spherical  polygons,  on  the  same  or  equal  spheres, 
are  Symmetrical  when  the  sides  and  angles  of  one  are  equal,  re- 
spectively, to  the  sides  and  angles  of  the  other,  if  the  equal 
parts  occur  in  opposite  orders. 

Thus,  if  spherical  ^ABC  and  A'B'C\  on 
the  same  or  equal  spheres,  have  sides  AB^  BC, 
and  CA  equal,  respectively,  to  sides  A'B'^  B'C\   ^^  ^ 

and  C'A',  and  A  A,  B,  and  C  to^  A',  B',  and  C",  and  the  equal  parts 
occur  in  the  opposite  orders  the  A  are  symmetrical. 


Proposition  XIII.     Theorem 

682.     TJie  spherical  triangles  corresponding  to  a  pair 
of  vertical  triedral  angles  are  symmetrical. 


Hypothesis.  AOA',  BOB',  and  COC  are  diameters  of  the 
sphere  with  center  0;  the  planes  determined  by  them  inter- 
sect the  spherical  surface  in®  ABA' B',  AC  A' C,  and  BCB'C. 

Conclusion.     Spherical  A  ABC  and  A'B'C  are  symmetrical. 

Suggestions.  — 1.   Prove  A' B'  =  AB ;  B'C  =  BC,  etc. 

2.  Prove  Z  BCA  =  L  B'C'A' ;  Z  BAG  =  Z  B'A'C ;  etc. 

3.  Prove  that  the  parts  of  A  ABC  occur  in  C' 

opposite  order  to  those  of  A  A'B'C .  y'"'^^/'    /*^\ 

The  adjoining  figure  will  aid  in  doing         h\^,jL..A^,\B'\ 

this.     A'B'C  has  been  slid  around  the  ^  ^J!^/_.__ir::ir.llVJ^' 

sphere  until  it  occupies  the  position  in-         \l7h~~~/- —  / 

dicated  in  this  figure.     Determine  the          \  l      /  / 

direction  from  A  to  B  to  C ;   and  also            \\  /  y^ 

the  direction  from  A'  to  B*  to  C.  p" 


430  SOLID   GEOMETRY  — BOOK  IX 

Pkoposition  XIY.     Theorem 

683.    Two  symmetrical  spherical  triangles,  of  lohich 
one  is  isosceles,  are  congruent  and  hence  equal. 


1 


Hypothesis.     A  ABC  is  symmetrical  to  AA'B^C^-,  that  is, 

AB  =  A^,  AC  =  A^y  BC=B^,  ZB  =  ZB',  ZC==ZG', 
Z  A=  ZA',  with  the  equal  parts  arranged  in  opposite  orders  in 

the  triangles  ;  also  AB  =  AC. 

Conclusion.  A  ABC  ^  A  A'B'C 

Proof.     1.   Since      AB  =  A^,  and  AB=  AC, 

.-.  A^=  AG. 

2.   In  like  manner  A^C  =  AB. 

Complete  the  proof  by  superposing  AA'B'C  on  A  ABC, 
making  A'C  coincide  with  AB,  with  point  A'  on  point  A. 

684.  Cor.  If  one  of  two  symmetrical  spherical  triangles  is 
isosceles,  the  other  is  also. 

Proposition  XV.     Theorem 

685.  Two  spherical  triangles  corresponding  to  a  pair 
of  vertical  triedral  angles  are  equal. 

Hypothesis.  AOA',  BOB',  and  COC  are  diameters  of  sphere 
0;  also,  the  planes  determined  by  them  intersect  the  surface 
in  arcs  AB,  BC,  AC,  A'B',  B'C,  and  A'C.     (Fig.  p.  431.) 


MEASUREMENT   OF  SPHERICAL  POLYGONS     431 
Conclusion.     Area  of  A  ABC  =  area  of  triangle  A'B'C. 

A 


Proof.  1.  Let  P  be  the  pole  of  the  small  circle  passing  through 
points  Aj  B,  and  C ;  draw  arcs  of  great  circles  PA,  PB,  and  PC. 

2.  ..PA  =  PB=PC.  §643 

3.  Draw  PP\  a  diameter  of  the  sphere,  and  P'A',  P'B',  and 
PC  arcs  of  great  CD ;  then  spherical  A  PAB  and  P'A'B'  are 
symmetrical.  §  682 

4.  But  spherical  A  PAB  is  isosceles. 

5.  .-.  A  PAB  =  A  P'A'B',  Why  ? 

6.  In  like  manner, 

A  PBC  =  A  P'B'C  and  A  PC  A  =  A  PC  A'. 

7.  Then  the  sum  of  the  areas  of  A  PAB,  PBC,  PAC  equals 
the  sum  of  the  areas  of  A  P'A'B',  P'B'C,  and  P'CA'. 

8.  .-.  area  A  ABC  =  area  A  A'B'C. 

686.  Cor.  Two  symmetrical  triangles  on  the  same  or  equal 
spheres  are  equal. 

1.  ljQtAA"B"C"  be  symmetrical  to /S  ABC',  let  A  A'B'C 
be  the  spherical  triangle  on  the  same  sphere  as  A  ABC,  such 
that  A',  B',  and  C  are  diametrically  opposite  to  A,  B,  and  C, 
respectively. 

2.  Then  A  A'B'C  is  also  symmetrical  to  A  ABC,  and  equal 
to  A  ABC. 

3.  .-.  the  parts  of  A  A"B"C"  and  A  A'B'C  are  equal  and  are 
arranged  in  the  same  order.     Hence  A  A"B"C"  ^  A'BC". 

4.  .'.AABC==AA"B"C". 


432  SOLID  GEOMETRY  —  BOOK  IX 

Proposition  XVI.     Theorem 

687.  A  spherical  triangle  of  a  sphere  equals  one  half  j 
a  lune  of  that  sphere  whose  angle  in  degrees  equals  the  1 
spherical  excess  of  the  triangle. 


Hypothesis.  A,  B,  and  C  are  the  measures  in  degrees  of  the 
angles  of  spherical  A  ABC.  E  represents  the  spherical  excess 
of  the  triangle. 

Conclusion.  A  ABC  =  \Le  ; 

that  is,  A  ABC  =  one  half  a  lune  whose  angle  is  E. 
Proof.   1.     Complete  the  ®  ABA'B',  BCB'C,  smd  AGA'C, 
and  draw  diameters  AA',  BB',  and  CC. 

2.  A  ABC  -^AACB'  =  lune  of  Z  B,  or  Lb- 

3.  A  ABC  -irAA'CB  =  lune  of  Z  A,  or  Lj,. 

4.  A  ABC  +  A  ABC  =  lune  of  Z  (7,  or  Lc. 

But  A  A'B'C  =  A  ABC,  so  §  685 

5.  A  ABC  -f-  A  A'B'C  =  lune  of  Z  C,  or  X^. 

6.  Adding  the  equations  of  steps  2,  3,  and  5, 

2  A  ^5(7  +  (A  ABC  +  A  ACB'  ^AA'CB-\-A  A'B'C) 

7.  But  A  ^50+ A  ^05' +  A  J[' 05  + A  ^'^'0 

=  surface  of  hemisphere 

=  Aso-  §  6^^ 

8.  .-.  2  A  ABC-]-  Li8o  =  i^^+^+c. 

9.  .*.  2  A  ^5(7  =i^+5+c'  —  Aso  =-^  (^+5+0-180' 

10.  But  J  +  5  +  (7-  180  =  E. 

11.  .-.  2  A  yl^C  =  X^,  or  A  ABC  =  i  L^. 


i 


MEASUREMENT   OF  SPHERICAL  POLYGONS     433 

688.  Cor.  1.  If  the  radius  of  the  sphere  is  Rand  the  sjjherical 
excess  of  A  ABC  in  degrees  is  E,  then 

689.  Cor.  2.     The  area  of  any  spherical  polygon  whose  excess 

IS  E  ts 

180 

Suggestion,  —  Divide  the  polygon  into  A  by  drawing  diagonals  from  one 
vertex.  Express  the  area  of  each  triangle.  Remember  that  the  excess  of  the 
polygon  equals  the  sum  of  the  excesses  of  the  triangles,  when  the  polygon  is 
divided  as  suggested. 

Note.  —  A  spherical  degree  may  be  defined  as  being  a  bi-rectangular 
spherical  triangle  whose  third  angle  is  one  spherical  angular  degree.  The 
area  of  a  spherical  triangle  in  spherical  degrees  can  be  proved  to  equal  its 
spherical  excess  in  degrees. 

Ex.  59.  Determine  the  area  of  a  spherical  triangle  whose  angles  are 
125°,  133°,  and  156°,  on  a  sphere  whose  radius  is  10  in. 

Ex.  60.  What  is  the  ratio  of  the  areas  of- two  spherical  triangles  on  the 
same  sphere  whose  angles  are  94°,  135°,  and  146°,  and  87°,  105°,  and  118°, 
respectively. 

Ex.  61.  Determine  the  area  of  a  spherical  triangle  whose  angles  are 
103°,  112°,  and  127°  on  a  sphere  whose  area  is  160. 

Ex,  62.  Find  the  area  of  a  spherical  hexagon  whose  angles  are  120°, 
139°,  148°,  155°,  162°,  and  167°,  on  a  sphere  whose  radius  is  12. 

Ex.  63.  The  sides  of  a  spherical  triangle  on  a  sphere  whose  radius  is 
15  in.  are  44°,  63°,  and  97°.     Find  the  area  of  its  polar  triangle. 

Ex.  64.  Determine  the  part  of  the  area  of  the  surface  of  a  sphere  in- 
tercepted by  a  triedral  angle  whose  face  angles  are  89°,  55°,  and  100°. 

Ex.  65.  Express  the  ratio  of  a  spherical  triangle  to  the  surface  of  the 
sphere  in  terms  of  the  spherical  excess  E  of  the  triangle,  when  (a)  the 
excess  is  measured  in  degrees  ;  (b)  the  excess  is  measured  in  right  angles. 

Ex.  66.  The  area  of  a  spherical  pentagon,  four  of  whose  angles  are 
112°,  131°,  138°,  and  168°,  is  27.  If  the  area  of  the  surface  of  the  sphere 
is  120,  what  is  the  other  angle  ? 

Ex.  67.  What  part  of  the  surface  of  a  sphere  is  a  tri-rectangular 
triangle  of  the  sphere  ? 

Ex.  68.  Compare  the  area  of  a  tri-rectangular  spherical  triangle  of  a 
sphere  whose  radius  is  10  in.  with  the  area  of  the  plane  triangle  formed  by 
the  chords  of  the  sides  of  the  spherical  triangle. 


434  SOLID   GEOMETRY  —  BOOK  IX 

VOLUME   OF  A  SPHERE 

690.  If  a  semicircle  be  revolved  about  its  diameter  as  an 
axis,  the  solid  generated  by  any  sector  of  the  semicircle  is 
called  a  Spherical  Sector. 

A 

Thus  if  semicircle  ACDB  be  revolved  about 
diameter  AB  as  an  axis,  sector  OCD  generates  a 
spherical  sector. 

The  zone  generated  by  the  arc  of  the  circu- 
lar sector  is  called  the  base  of  the  spherical 
sector. 

Note.  — In  the  following  pages,  the  expression  "Vol.  OCD'^  will  be 
used  to  denote  the  volume  of  the  solid  generated  by  revolving  the  portion 
of  the  plane  within  OCD  around  some  axis  specified. 

691.  Application  of  Limits  to  Spherical  Sectors.    Let  O  be 

the  center  of  arc  AB,  and  OM  be  any  diameter  of  the  circle 
whose  center  is  O.     Let  G  bisect  arc  AB,  and 
draw  radii  OA,     OB,  and  00.     Draw  OE  ± 
AC;  draw  broken  line  ACB. 

If  the   sector  OAB  is  revolved  about  OM 
as   an  axis,   it  generates  a  spherical  sector. 
The  portion  of  the  plane  bounded  by  polygon 
OACB  generates  a  solid  which  is  less  than  the  spherical  sector 
generated  by  circular  sector  OAB. 

If  arcs  AG  and  GB  are  bisected  at  D  and  F  respectively 
and  broken  line  ADGFB  is  drawn,  then  when  the  figure  is 
revolved  about  OM,  the  part  of  the  plane  bounded  by  polygon 
OADGFB  generates  a  solid  more  nearly  equal  to  the  spherical 
sector.  If  the  process  of  halving  the  arcs  be  continued  indefi- 
nitely, it  will  be  assumed  evident  that  the  solid  generated  by 
the  part  of  the  plane  bounded  by  the  inscribed  polygon  ap- 
proaches the  spherical  sector  as  limit. 

Notice  that  the  surface  generated  by  broken  line  ADGFB 
approaches  as  limit  the  zone  generated  by  arc  AGB.     (§  668.) 


VOLUME  OF  A  SPHERE 


435 


Proposition  XVII.     Theorem 

692.  If  a  portion  of  a  plane  inclosed  hy  an  isosceles 
triangle  he  revolved  about  a  straight  line  in  its  plane 
as  axis,  ivhich  passes  through  its  vertex  ivithout  inter- 
secting its  surface  and  ivithout  being  parallel  to  its 
base,  the  volume  of  the  solid  generated  is  equal  to  the 
area  of  the  surface  generated  by  its  base  multiplied  by 
one  third  its  altitude. 


Hypothesis.  Isosceles  A  OAB,  and  the  surface  inclosed,  are 
revolved  about  straight  line  OF  in  its  plane  ;  OF  is  not  II  base 
AB;  OC±AB. 

Conclusion.     Vol.  OAB  =  area  AB  x  ^  00. 

Proof.  1.  Draw  AD  ±  OF  and  BE  A.  OF;  extend  BA  to 
meet  OF  at  F. 

2.   Vol.  OBF=  vol.  OBE  -f  vol.  BEF 

=  \TrB^  X  0^  +  i ttBE"  X  EF  §  616 

=  \7rBE'{0E  4-  EF)=^7rBE  X  BE  x  OF. 
BE  X  OF  =  OCX  BF.  Prove  it. 


3. 
4. 

5. 
6. 
7. 
8. 
9. 
10. 

11. 


But 


But 


•.  vol.  OBF=  \  ttBE  X  OC  X  BF. 


area  BF  x  ^  OC. 
vol.  OAF  =  area  AF  x  ^  OC. 


ttBE  X  BF  is  the  area  BF. 
.;.  vol.  OBF 
Similarly 
Subtracting  9  from  8, 

vol.  OAB  =  (area  BF  -  area  AF)  x  ^  OC. 
=  area  AB  X  ^  OC. 


§614 


436  SOLID   GEOMETRY  —  BOOK  IX 

Pkoposition  XYIII.     Theorem 

693.  The  volume  of  a  spherical  sector  is  equal  to  the 
area  of  the  zone  which  forms  its  base,  multiplied  hy 
one  third  the  radius  of  the  sphere. 


Hypothesis.  Sector  OAB  of  O  0  is  revolved  about  di- 
ameter OM  as  an  axis  ;  E  is  the  radius  of  the  sphere. 

Conclusion.  Vol.  of  the  spherical  sector  generated  by  cir- 
cular sector  OAB  =  area  of  zone  generated  by  AB  X  ^B. 

Proof.  1.  Let  C bisect^.  Draw  AC,  CB,  OC \  and  draw 
OEl.Aa 

2.  Vol.  OAC  =  area  AG  x  \  OE.  §  691 

3.  Vol.  OCB  =  area  CB  x  \  OE. 

4.  Adding,  vol.  0^05  =  (area  AC  +  area  CB)  x  \  OE 

=  area  ACB  x  \  OE. 

5.  Let  the  subdivisions  of  AB  be  bisected  indefinitely. 

6.  Then  vol.  OACB  =  vol.  generated  by  sector  OAB, 
and  area  ACB  =  area  of  zone  generated  by  AB 
and  OE  =  R. 

7.  .*.  vol.  generated  by  sector  OAB 

=  area  of  zone  generated  by  AB  x  ^B.  §  403 

694.  Cor.  1.  If  V  denotes  the  volume  of  a  spherical  sector,  h 
the  altitude  of  the  zone  which  forms  its  base,  and  R  the  radius  of 
the  sphere, 

V  =  27rRh  xiR  =  lirR'h.  §  671 

695.  Cor.  2.  The  sphere  may  be  regarded  as  a  spherical 
sector  whose  base  is  the  entire  surface  of  the  sphere.     Letting 


VOLUME  OF  A  SPHERE  437 

V  denote  the  volume  of  the  sphere,  and  M  its  radius, 

The  volume  of  a  sphere  is  equal  to  the  cube  of  its  radius  multi- 
plied by  j^TT. 

696.  Cor.  3.  The  volumes  of  two  spheres  have  the  same  ratio 
as  the  cubes  of  their  ludii. 

Ex.  69.     Find  the  volume  of  the  sphere  whose  radius  is  12. 

Ex.  70.  Determine  the  volume  of  metal  in  a  spherical  shell  10  in.  in 
diameter  and  1  in.  thick. 

Ex.  71.  A  spherical  cannon  ball  9  in.  in  diameter  is  dropped  into  a 
cubical  box  filled  with  water,  whose  depth  is  9  in.  How  many  cubic 
inclies  of  water  will  be  left  in  the  box  ? 

Ex.  72.  If  a  sphere  0  in.  in  diameter  weighs  351  oz.,  what  is  the 
weight  of  a  sphere  of  the  same  material  whose  diameter  is  10  in.? 

Ex.  73.  The  outer  diameter  of  a  spherical  shell  is  9  in.,  and  its  thick- 
ness is  1  in.  What  is  the  weight,  if  a  cubic  inch  of  the  metal  weighs  one 
third  pound  ? 

Ex.  74.  Find  the  area  of  the  surface  and  the  volume  of  the  sphere 
inscribed  in  a  cube  the  area  of  whose  surface  is  486  sq.  in. 

Ex.  75.  Find  the  radius  and  the  volume  of  a  sphere,  the  area  of 
whose  surface  is  324  ir  sq.  in. 

Ex.  76.  Prove  that  the  volume  of  a  sphere  is  two  thirds  the  volume 
of  its  circumscribed  cylinder. 

Ex.  77.  "Within  a  sphere  of  radius  E  is  inscribed  a  right  circular 
cylinder  whose  altitude  equals  the  diameter  of  its  base. 

(a)  Determine  its  lateral  area  and  compare  the  result  with  the  area  of 
the  surface  of  the  sphere. 

(b)  Compare  its  volume  with  the  volume  of  the  sphere. 

Ex.  78.  Given  a  spherical  surface  of  radius  R  and  its  circumscribed 
right  circular  cylinder.  From  the  center  of  the  sphere,  draw  lines  to  the 
points  of  the  circles  bounding  the  bases  of  the  cylinder,  thus  forming  the 
two  right  circular  cones.  Compare  the  volume  of  the  sphere  with  the  dif- 
ference between  the  volume  of  the  cylinder  and  the  sum  of  the  volumes 
of  the  two  cones. 

Ex.  79.  A  cylindrical  vessel,  8  in.  in  diameter,  is  filled  to  the  brim 
with  water.  A  ball  is  immersed  in  it,  displacing  water  to  the  depth  of  2  J 
in.     Find  the  diameter  of  the  ball. 

Note.  —  Supplementary  Exercises  72-85,  p.  460,  can  be  studied  now. 


438  SOLID  GEOMETRY  —  BOOK  IX 

SUPPLEMENTARY  TOPICS 

Group  A.     Construction  of  Spheres 

Proposition  XIX.     Theorem 

697.    Through  four  points  not  in  the  same  plane,  one  and  only 
owe  spherical  surface  can  be  passed. 


Hypothesis.  A,  B,  O,  and  D  are  four  points,  not  in  the  same 
plane. 

Conclusion.  One  and  only  one  spherical  surface  can  be  passed 
through  A,  B,  C,  and  Z>. 

Proof.  1.  Every  point  equidistant  from  O  and  D  must  lie 
in  a  plane  EKFl.  CD  at  its  mid-point  K\  and  conversely. 

§§457;  459 

2.  Every  point  equidistant  from  B  and  G  must  lie  in  a  plane 
IJFJ^BG  2it  its  mid-point  J-,  and  conversely. 

3.  These  planes  intersect  in  a  line  HF,  which  is  ±  plane 
BCD  at  F,  the  circumcenter  of  A  BCD.  §  499 

Also,  by  steps  1  and  2,  every  point  equidistant  from  B,  G, 
and  D  must  lie  in  HF-,  and  conversely. 

4.  Similarly,  every  point  equidistant  from  A,  G,  and  D  lies 
in  line  OE,  which  is  ±  plane  ACD  at  E,  the  circumcenter  of 
AACD. 

5.  Since  E  and  F  are  in  plane  EKF,  then  OE  and  HF  lie 
in  plane  EKF.  §  497 


CONSTRUCTION  OF  SPHERES  439 

6.  .-.  OE  intersects  HF  at  a  point  0. 

7.  .'.  0  is  one  and  the  only  point  which  is  equidistant  from 
A,  B,  C,  and  D.  Steps  3  and  4 

8.  .-.a  sphere  with  center  0  and  radius  OA  is  the  one  and 
only  sphere  through  Af  B,  C,  and  D. 

698.  Cor.  A  sphere  may  he  circumscribed  about  any  tetror 
edron. 

Ex.  80.  What  is  the  locus  of  the  center  of  a  sphere  which  will  have 
a  given  radius  r  and  will  pass  through  a  given  point  P  ? 

Ex.  81.  What  is  the  locus  of  the  center  of  a  sphere  which  will  pass 
through  each  of  two  given  points  ? 

Ex.  82.  What  is  the  locus  of  the  center  of  a  sphere  which  will  pass 
through  each  of  three  given  points  ? 

Ex.  83.  What  is  the  locus  of  the  center  of  a  sphere  which  will  pass 
through  all  the  points  of  a  given  circle  ? 

Ex.  84.  Is  it  possible  for  a  sphere  to  pass  through  all  the  points  of  a 
given  circle  and  also  through  a  given  point  outside  the  plane  of  the  circle  ? 
If  so,  tell  how  to  determiue  its  center  and  its  radius. 

Ex.  85.     Prove  that  a  sphere  can  be  circumscribed  about  a  cube. 

Ex.  86.  What  is  the  locus  of  the  center  of  a  sphere  which  is  tangent 
to  a  given  plane  at  a  given  point  ? 

Ex.  87.  What  is  the  locus  of  the  center  of  a  sphere  which  will  have 
a  given  radius  r  and  will  be  tangent  to  a  given  plane  ? 

Ex.  88.  What  is  the  locus  of  the  center  of  a  sphere  which  will  be 
tangent  to  each  of  the  faces  of  a  given  diedral  angle  ? 

Ex.  89.  Is  it  possible  for  a  sphere  to  be  tangent  to  each  of  the  faces 
of  a  given  triedral  angle  ?  If  there  is  more  than  one  such  sphere,  what 
is  the  locus  of  the  center  ? 

Ex.  90.  Find  the  area  of  the  spherical  surface 
passing  through  the  vertices  of  a  regular  tetraedron 
whose  edge  is  8. 

Suggestion.  —  Draw  DOE  and  AOF  perpendicular 
to  &^  ABC  and  BCD  respectively. 

C 
Ex.  91.    Prove  that  a  sphere  can  be  inscribed  in  a  given  tetraedron. 


440  solid  geometry  —  book  ix 

Group  B.    General  Theorems  of  Spherical 
Geometry 

699.  Special  interest  attaches  to  the  following  theorems 
because  of  their  similarity  to  certain  theorems  of  plane  geom- 
etry. In  each  case,  the  pupil  should  recall  the  corresponding 
theorem  of  plane  geometry,  if  there  is  one,  or  should  note  the 
difference  between  the  theorem  of  spherical  geometry  and  the 
corresponding  situation  in  plane  geometry. 

Proposition  XX.     Theorem 

700.  The  intersection  of  two  spherical  surfaces  is  a  circle,  whose 
center  is  in  a  straight  line  joining  the  centers  of  the  spheres  and 
whose  plane  is  perpendicular  to  that  line. 


Hypothesis.     0  and  0'  are  the  centers  of  two  intersecting 

spherical  surfaces. 

Conclusion.  The  intersection  of  the  surfaces  is  a  circle 
whose  center  is  in  line  00'  and  whose  plane  is  A.  00'. 

Proof.  1.  Through  0  and  0'  and  any  point  A  of  the  inter- 
section, pass  a  plane.  This  plane  cuts  the  two  surfaces  in  two 
intersecting  great  (D.  Let  AB  be  the  common  chord  of  these 
two  (D,  intersecting  00'  at  C. 

2.  .'.  00'  bisects  AB  at  right  angles.  §  207 

3.  If  the  entire  figure  is  revolved  about  00'  as  an  axis,  the 
(D  will  generate  the  spherical  surfaces  whose  centers  are  0 
and  0'. 

Point  A  will  generate  a  O  whose  center  is  C  and  radius 
AC,  which  is  common  to  the  two  spherical  surfaces. 

4.  The  plane  of  O  ^C  is  ±  00'.  §  458 


THEOREMS  OF  SPHERICAL  GEOMETRY         441 

6.  No  point  outside  O  ACB  can  lie  in  both  surfaces ;  for,  if 
there  were  such  a  point,  the  two  surfaces  would  necessarily 
coincide.  §  697 

Ex.  92.  What  is  the  locus  of  points  at  the  distance  ri  from  a  given 
point  Pi  and  at  the  distance  rg  from  a  given  point  Pa  ? 

Ez.  93.  The  distance  betveeen  the  centers  of  two  spheres  whose 
radii  are  25  and  17,  respectively,  is  28.  Find  the  diameter  of  their  circle 
of  intersection,  and  its  distance  from  the  center  of  each  sphere. 

Suggestion.  —  Recall  §  313. 

Proposition  XXI.     Theorem 

701.   Any  side  of  a  spherical  triangle  is  less  than  the  sum  of 

the  other  two. 

A 


Hypothesis.     AB  is  any  side  of  spherical  A  ABO. 
Conclusion.  AB  <  AC  4-  BC. 

Suggestions.  — ■[.  Compare  /L  AOB  with  A  AOC+  ABOC. 
2.  What  is  the  relation  of  the  measure  of  AB  and  that  of  /L  AOB'i    Of 
iC'and  AA0C1  etc. 

Ex.  94.  Prove  that  any  side  of  a  convex  spherical  pojygon  is  less 
than  the  sum  of  the  remaining  sides. 

Ex.  95.  Prove  that  the  sum  of  the  arcs  of  great  circles  drawn  from 
any  point  within  a  spherical  triangle  to  the  extremities  of  any  side,  is  less 
than  the  sum  of  the  other  two  sides  of  the  triangle. 

702.  Two  spherical  polygons  on  the  same  sphere  or  equal 
spheres  are  mutually  equilateral  or  mutually  equiangular  when 
the  sides  or  angles  of  one  are  equal  respectively  to  the  sides  or 
angles  of  the  other,  whether  taken  in  the  same  or  in  opposite 
orders. 


442  SOLID   GEOMETRY  —  BOOK   IX 

Proposition  XXII.     Theorem 

703.  If  two  spherical  triangles  on  the  same  sphere,  or  equal 
spheres,  have  two  sides  and  the  included  angle  of  one  equal  respec- 
tively to  two  sides  and  the  included  angle  of  the  other, 

I.  They  are  congruent  if  the  equal  parts  occ^ir  in  the  same 
order. 

II.  They  are  symmetrical  if  the  equal  parts  occur  in  opposite 
orders. 


1.  Hypothesis.  ABC  and  DEF  are  spherical  A  on  the  same 
sphere,  or  equal  spheres,  having  AB  =  DE,  AC  =  DF,  and 
/.A=/.Dy,  and  the  equal  parts  occur  in  the  same  order. 

Conclusion.  A  ABC  ^  A  DEF. 

Suggestion.  —  Prove  it  by  superposition  as  in  §  63. 

II.  Hypothesis.  ABC  and  D'E'i'  are  spherical  A  on  the 
same  sphere,  or  equal  spheres,  having  AB  =  D'E\  AC  =  D'F', 
and  AA  =  ZD';  and  the  equal  parts  occur  in  opposite  orders. 

Conclusion.     ABC  and  D'E'F'  are  symmetrical  A. 

Proof.  1.  Let  DEF  be  a  spherical  A  on  the  same  sphere, 
or  an  equal  sphere,  symmetrical  to  AD'E'F',  having 

DE  =  UE\     DF=  UF',  and  Z  Z>  =  Z  D', 
the  equal  parts  occurring  in  opposite  orders. 

2.  Then  in  spherical  A  ABC  and  DEF, 

AB  =  DE,  ACJ  =  DF,  and  Z  A  =  Z  D; 
and  the  equal  parts  occur  in  the  same  order. 

3.  .-.  A  ABC  ^  A  DEF. 

4.  .-.  A  ABC  is  symmetrical  to  A  D'E'F'. 


THEOREMS  OF  SPHERICAL  GEOMETRY         443 

Ex.  96.  Prove  that  the  arc  of  a  great  circle  bisecting  the  vertical 
angle  of  an  isosceles  spherical  triangle  is  perpendicular  to  the  base  and 
bisects  the  base, 

Ex.  97.  Prove  that  the  angles  opposite  the  equal  sides  of  an  isos- 
celes spherical  triangle  are  equal. 

Proposition   XXIII.     Theorem 

704.  If  two  spherical  triangles  on  the  same  sphere^  or  on  equal 
spheres  J  have  a  side  and  two  adjacent  angles  of  one  equal  respec- 
tively to  a  side  and  two  adjacent  angles  of  the  other, 

I.  They  are  congruent  if  the  equal  parts  occur  in  the  same  order. 

II.  They  are  symmetrical  if  the  equal  parts  occur  in  opposite 
orders. 

The  proof  is  left  to  the  student.  • 

Proposition  XXIV.     Theorem 

705.  If  two  spherical  triangles  on  the  same  sphere,  or  on  equal 
spheres,  are  mutually  equilateral,  they  are  mutually  equiangular. 


-"^0 


Hypothesis.      ABC   and    DEF   are    mutually    equilateral 

spherical  A  on  the  equal  sphere  ;  BC  and  EF  are  homologous. 

Conclusion.     A  ABC  and  DEF  are  mutually  equiangular. 

Suf/f/estioiis.  —  1,  Let  O  and  0'  be  the  centers  of  the  respective  spheres,  and 
draw  the  radii  to  A,B,  C,  D,  E,  and  F.  Consider  the  two  triedral  angles 
O- ABC  And  O'-DEF. 

2.  Compare  the  face  angles  of  triedrals  0-ABC  and  O'-DEF. 

3.  Compare  the  diedral  angles  of  O-ABC  and  O'-DEF. 

4.  Now  compare  the  A  A  and  D  ;  also  the  A  B  and  /;  ;  also  A  C  and  F. 
Note. — The  theorem  may  be  proved  in  a  sinnlar  manner  when  the 

given  spherical  ^  are  on  the  same  sphere. 


444  SOLID   GEOMETRY  —  BOOK  IX 

706.  Cor,  If  two  spherical  triajigles  on  the  same  sjjhere,  or  on 
equal  spheres,  are  mutually  equilateral, 

I.  They  are  congruent  if  the  equal  parts  occur  in  the  same 
order. 

II.  Tliey  are  symmetrical  if  the  equal  parts  occur  in  opposite 
orders. 

Ex.  98.  Prove  that  the  arc  of  a  great  circle  drawn  from  the  vertex 
of  an  isosceles  spherical  triangle  to  the  middle  point  of  the  base,  is  per- 
pendicular to  the  base,  and  bisects  the  vertical  angle. 

Proposition  XXV.     Theorem 

707.  If  two  spherical  triangles  on  the  same  sphere,  or  equal 
spheres,  are  mutually  equiaiigular,  their  polar  triangles  are 
mutually  equilateral. 


i 


Hypothesis.  ABC  and  DEF  are  mutually  equiangular 
spherical  A  on  the  same  sphere  or  equal  spheres,  A  A  and 
D  being  homologous  ;  also,  A  A'B'C  is  the  polar  A  of  A  ABC, 
and  A  D'E'F'  of  A  DEF,  A  being  the  pole  of  mj',  and  D  of 
E'F'. 

Conclusion.     A  A'B'C  and  D'E'F'  are  mutually  equilateral. 

Suggestions.  — 1.  Compare  the  measure  ot  Z.  A  and  B'C";  oi  Z.  D  and 
E^'.    Then  compare  B^'  and  E^'. 

2.  Proceed  similarly  for  the  other  pairs  of  homologous  sides. 

708.  Cor.  If  two  spherical  triangles  on  the  same  sphere,  or 
equal  spheres,  are  mutually  equilateral,  their  polar  triangles  are 
mutually  equiangular. 

Suggestion.  —  Model  the  proof  after  that  of  §  707. 


THEOREMS  OF  SPHERICAL  GEOMETRY        445 

Proposition  XXVI.     Theorem 

709.   If  two  sjjherical  triangles  on  the  same  sphere,  or  on  equal 
spheres,  are  mutually  equiangular,  they  are  mutually  equilateral. 


D' 


Hypothesis.  ABC  and  DEF  are  mutually  equiangular  spher- 
ical A  on  the  same  sphere  or  equal  spheres. 

Conclusion.     A  ABC  and  DEF  are  mutually  equilateral. 

Proof.  1.  Let  A  A'B'C  be  the  polar  A  of  ABC,  and  D'E'F 
of  DEF. 

2.  Since  A  ABC  and  A  DEF  are  mutually  equiangular, 
A  A'B'C  and  A  D'E'F'  are  mutually  e(iuilateral.  §  707 

3.  .'.  A  A'B'C  and  A  D'E'F'  are  mutually  equiangular. 

4.  .-.  But  A  ABC  is  the  polar  A  of  A'B'C  and  DEF  of 
D'E'F.  §  655 

5.  .*.  A  ABC  and  A  DEF  are  mutually  equilateral.       Why  ? 

710.  Cor.  Iftv:o  spherical  triangles  on  the  same  sphere  or  on 
equal  spheres  are  mutually  equiangular, 

I.  They  are  congruent  if  the  equal  parts  are  arranged  in  the 
same  order. 

II.  They  are  symmetrical  if  the  equal  paHs  are  arranged  in 
opposite  orders. 

Ex.  99.  Compare  the  theorem  of  Proposition  XXVI  with  the  corre- 
sponding theorem  about  two  plane  triangles. 

Ex.  100.  If  three  diameters  of  a  sphere  be  drawn  so  that  each  is  per- 
pendicular to  the  other  two,  the  planes  determined  by  them  divide  the 
surface  of  the  sphere  into  eight  congruent  tri-rectangular  triangles. 

Note.  —  Recall  at  this  point  all  the  theorems  by  which  two  spherical 
triangles  can  be  proved  congruent. 


446 


SOLID   GEOMETRY  —  BOOK   IX 


Proposition  XXVII.     Theorem 

711.   In  an  isosceles  triangle,  the  angles  opposite  the  equal  sides 
are  equal. 


D   *C 

Hypothesis.     In  spherical  triangle  ABC,  AB  =  AG. 
Conclusion.  ZB=  ZC. 

Suggestion.  —  Let  AD,  an  arc  of  a  great  circle,  bisect  BC    Use  §  705. 


Proposition  XXYIII.     Theorem 

712.   If  two  angles  of  a  spherical  triangle  are  equal,  the  sides 
opposite  are  equal. 


Hypothesis.     In  spherical  A  ABG,  Z.B=  Z.C. 
Conclusion.  AB  =  AG. 

Suggestions.  —  1.  Let  A  A'B'C  be  the  polar  A  of  ABC,  B  being  the  pole  of 
iX",and  CoiA^'. 

2.  Compare  A'C  and  A'B'  by  using  §  656. 

3.  Compare  Z  B'  and  Z  C,  and  from  them  determine  the  relation  between 
/b  and  Id. 

Ex.  101.  Prove  that  the  great  circle  arcs  drawn  to  the  extremities  of 
an  arc  of  a  great  circle  from  any  point  on  the  great  circle  perpendicular 
to  and  bisecting  it  are  equal. 

Ex.  102.     State  and  prove  the  converse  of  Ex.  101. 


THEOREMS  OF  SPHERICAL  GEOMETRY         447 

Proposition  XXIX.     Theorem 

713.  If  tivo  angles  of  a  spherical  triangle  are  unequal,  the  sides 
opposite  them  are  unequal,  the  side  (ypposite  the  greater  angle  being 
the  greaier. 


.'.BD  =  Da 

Why? 

Ab^Bb>AB. 

Why? 

\Ab-\-DC>AB. 

.\AC>AB. 

Hypothesis.     In  spherical  A  ABC,  Z  ABC  >  Z  C 

Conclusion.  AC  >  AB. 

Proof.  1.  Let  BD,  an  arc  of  a  great  circle,  meeting  AC  at 
D,  make  Z  CBD  =ZC 

2.  ^         ^ 

3. 
4. 
5. 

714.  Cor.  If  turn  sides  of  a  spherical  tnangle  are  unequal, 
the  angles  opposite  are  unequal,  the  angle  opposite  the  greater  side 
being  the  greater. 

An  indirect  proof  based  upon  §§  712  and  713  is  quite  easy. 

Bac.  103.  What  is  the  locus  of  points  on  the  surface  of  a  sphere  which 
are  equidistant  from  the  extremities  of  an  arc  of  a  great  circle  of  that 
sphere  ? 

Ex.  104.  Prove  that  the  great  circle  arcs  perpendicular  to  and 
hisecting  the  sides  of  a  spherical  triangle  intersect  in  a  point  which  is 
equidistant  from  the  vertices  of  the  triangle. 

Ex.  105.  Prove  that  a  circle  can  be  circumscribed  about  a  spherical 
triangle. 


448  SOLID   GEOMETRY  —  BOOK  IX 

Proposition  XXX.     Theorem 

715.  The  shortest  line  on  the  surface  of  a  sphere  between  tivo 
given  points  is  the  arc  of  a  great  circle,  not  greater  than  a  semi- 
circle which  joins  the  two  points. 


Hypothesis.     Points  A  and  B  are  on  the  surface  of  a  sphere, 
and  AB  is  an  arc  of  a  great  O,  not  greater  than  a  semicircle. 

Conclusion.     AB  is  the  shortest  line  on  the  surface  of  the 
sphere  between  A  and  B. 

Note.  —  The  following  proof  is  divided  into  four  parts,  (a),  (6),  (c), 
and  (d). 

Proof,     (a)   1.    Let  C  be  any  point  in  AB. 
2.    Let  DCF  and  ECG  be  arcs  of  small  (D  with  A  and  B 
respectively  as  poles,  and  AC  and  BC  as  polar  distances. 

(b)  DCF  and  ECG  have  only  point  C  in  common. 

1.  For  let  F  be  any  other  point  in  DCF  and  draw  AF  and 
BF,  arcs  of  great  circles. 

2.  .-.  i?=iC.  §643 

3.  But  AF-\-BF>AC  +  BC.  Why? 

4.  Subtracting  AF  from  the  first  member  of  the  inequality 
and  its  equal  AC  from  the  second  member, 

BF>BC,  or  BF>BG. 

5.  .-.  i?^lies  outside  small  QECG,  and  DCF  and  ECG  have 
only  point  C  in  common. 

(c)  The  shortest  line  on  the  surface  of  the  sphere  from  A  to 
B  must  pass  through  C. 


THEOREMS  OF  SPHERICAL  GEOMETRY         449 

1.  Let  ADEB  be  any  line  on  the  surface  of  the  sphere  be- 
tween A  and  B^  not  passing  through  C,  and  cutting  DCF  and 
ECG  at  D  and  E  respectively. 

2.  Then,  whatever  the  nature  of  line  AD,  it  is  evident  that 
an  equal  line  can  be  drawn  from  A  to  C. 

3.  In  like  manner,  whatever  the  nature  of  line  BE,  an  equal 
line  can  be  drawn  from  B  to  C. 

4.  Hence  a  line  can  be  drawn  from  AtoB  passing  through 
O,  equal  to  the  sum  of  lines  AD  and  BE,  and  consequently 
less  than  ADEB  by  the  part  DE. 

5.  Therefore  no  line  which  does  not  pass  through  C  can  be 
the  shortest  line  between  A  and  B. 

(d)  AB  is  the' shortest  line  from  ^  to  JB  on  the  surface  of 
the  sphere. 

1.  But  C  is  any  point  in  AB. 

2.  Hence  the  shortest  line  from  ^  to  i^  must  pass  through 
every  point  of  AB. 

3.  Then  the  great  circle  arc  AB  is  the  shortest  line  on  the 
surface  of  the  sphere  between  A  and  B. 

Ex.  118.  Prove  that  any  point  in  the  arc  of  a 
great  circle  bisecting  a  spherical  angle  is  equally 
distant  (§  573)  from  the  sides  of  the  angle. 

(Prove  PM  =  PN.     Let  ^  be  a  pole  of  arc  AB, 
and  F  of  arc  BC.     Spherical  A  BPE  and  BFF     ^^-' 
are  symmetrical  by  §  702,  XL,  and  PE  =  PF.) 

Ex.  119.     Prove  that  a  point  on  the  surface  of 
a  sphere,  equally  distant  from  the  sides  of  a  spherical  angle,  lies  in  the  arc 
of  a  great  circle  bisecting  the  angle. 

(Fig.  of  Ex.  118.  Prove  Z  ABP  =  Z  CBP.  Spherical  ^  BPE  and  BPF 
are  symmetrical  by  §  706. ) 

Ex.  120.  What  is  the  locus  of  points  on  the  surface  of  a  sphere 
equally  distant  from  the  sides  of  a  spherical  angle? 

Ex.  121.  Prove  that  the  arcs  of  great  circles  bisecting  the  angles  of 
a  spherical  triangb  meet  in  a  point  equally  distant  from  the  sides  of  the 
triangle. 

Ex.  122.  Prove  that  a  circle  may  be  inscribed  in  any  spherical  tri- 
angle. 


450 


SOLID   GEOMETRY  —  BOOK  IX 


Gkoup   C.     Spherical   Segments,   Pyramids, 
AND  Wedges 

716.  A  Spherical  Segment  is  the  portion  of  a  sphere  included 
between  two  parallel  planes  which  intersect  the  sphere. 

The  portions  of  the  planes  bounding  the  segment  are  the 
bases  of  the  segment ;  the  perpendicular  between  the  planes  is 
the  altitude  of  the  segment. 

A  spherical  segment  of  one  base  is  the  spherical  segment 
one  of  whose  bounding  planes  is  a  tangent  to  the  sphere. 

If  a  semicircle  ACEB  be  revolved  about 
diameter  AB  as  an  axis,  and  CD  and  EF  are 
perpendicular  to  AB,  the  portion  of  the 
plane  bounded  by  FECD  generates  a  spheri- 
cal segment  whose  altitude  is  DF,  and  whose 
bases  have  radii  CD  and  EF  respectively ; 
the  portion  ACD  generates  a  spherical  seg- 
ment of  one  base  wjiose  altitude  is  AD. 

lYJ.  If  r  and  r'  are  the  radii  of  the  bases,  h  the  altitude,  and 
V  the  volume  of  a  spherical  segment,  then 


Let  O  be  the  center  of  ADB ;  let  AA'  and  BB'  be  ±  to  the 
diameter  OM;  let  AA'  =  v' ,  BB'  =  r,  A'B'  ==  h.  Let  the  whole 
figure  revolve  about  OM  as  an  axis.  Let  v  =  the  volume  of 
the  resulting  spherical  segment. 

Solution.     1.   Draw  OA,  OB,  and  AB ;  draw  OC  ±  AB,  and 
AE  ±  BB'.     Let  OA  =  R. 


SPHERICAL  SEGMENTS  AND  WEDGES  451 

2.  Now,  vol.  ADBB'A'  =  vol.  ACBD  +  vol.  ABB' A'.     (1) 

3.  Also,       vol.  ACBD  =  vol.  OADB  -  vol.  OAB. 

4.  vol.  OADB  =  I  Tram.  §  694 

5.  And,  vol.  OAB  =  a^renAB  x  i  00  §  691 

6.  =hx2  ttOC  X  i  OC  §  668 

7.  =1  TrOO'/i. 

8.  .-.  vol.  ACDB  =  1 7r727i  -  |  ttOO'A 

9.  =  J  7r(i22  -  OC^)h. 

10.    But,  Ii^-OC^  =  AC^^  Why? 

11.  =  a^)2 

12.  =  \  Alf, 

13.  "      .'.\ol  ACDB  =  ^TrXiA&x  h  =  iwA&h. 

14.  Now,  Zb'  =  5^'  +  3E' 

15.  =(r-r')2+A2. 

16.  .-.  vol.  ^(7Z>i^  =  i  7r[(r  -  r')2  +  Ji^yi. 

17.  Also,      vol.  ABB' A'  =  J  7r(r2  +  r"'  -\-  rr')h.  §  621 

18.  Substituting  in  step  2,  vol.  ADBB'A' 

=  i  7r[(r  -  r'y  +  7i2]/i  +  I  ^(2  r2  +  2  r'2  +  2  rr')/* 

19.  =  J  7r(r2  -  2  rr'  +  r'2  -\- h^ -^  2  r^ -{- 2  r'^ -{- 2  rr')h 

20.  =  -J  7r(3  7-2  +  3  r'2)/4  4-  -^  nh^ 

21.  =i7r(r2+r'2)A  +  i7r/il 

Ex.  123.  Find  the  volume  of  a  spherical  segment,  the  radii  of  whose 
bases  are  4  and  5,  and  whose  altitude  is  9. 

718.  A  Spherical  Wedge  is  a  solid  bounded  by  a  lune  and 
the  planes  of  its  bounding  arcs. 

Evidently,  two  spherical  wedges  in  the  same  sphere,  or 
equal  spheres,  are  congruent  when  their  angles  are  equal. 

Also,  it  is  evident  that  two  wedges  in  the  same  sphere  can 
be  added  by  placing  them  so  that  they  have  one  common 
bounding  plane.  The  angle  of  the  sum  is  equal  to  the  sum  of 
the  angles  of  the  wedges. 

Note.  —  Review  at  this  time  §  675  and  §  676,  noting  the  analogy  be- 
tween wedges  and  lunes. 


452 


SOLID   GEOMETRY — BOOK  IX 


719.  It  can  be  proved  as  in  §  677  that  two  wedges  have  the 
same  ratio  as  their  angles.     (Cf.  §  677.) 

720.  A  sphere  may  be  regarded  as  a  wedge  whose  angle  is 
360°.     (Cf.  §  678.) 

Therefore,  a  wedge  of  a  sphere  whose  radius  is  r,  whose  angle 
contains  A  degrees  has  a  volume  v  determined  as  follows  : 


or  V  = 


TTT^A 


irr 


360'  270 

721.  Since  the  lune  whose  angle  is  A  degrees,  on  a  sphere 
whose  radius  is  r,  has  its  area  expressed  by  the  formula  ^^^-^ 

(§  680) 

.*.  the  volume  of  a  wedge  equals  one  third  the  radius  of  the 

sphere  multiplied  by  the  area  of  the  lune  which  forms  its  base. 

722.  A  Spherical  Pyramid  is  a  solid 
bounded  by  the  spherical  polygon  and  the 
planes  of  its  sides  ;  as  0-ABGD  in  the  ad- 
joining figure. 

The  center  of  the  sphere  is  the  vertex  of 
the  pyramid,  and  the  spherical  polygon  is 
its  base. 

Two    spherical    pyramids   are    congruent 
when  their  bases  are   congruent,   for   they   can   be    made    to 
coincide. 

723.  Two  spherical  pyramids  whose  bases  are  symmetrical 
isosceles  spherical  triangles  are  congruent,  for  their  bases  are 
congruent  by  §  683. 

724.  Two  spherical  pyramids  corre- 
sponding to  a  pair  of  vertical  triedral 
angles  are  equal.     (Cf.  §  685.) 

Suggestions.  —  1.   Kecall  the  proof  of  §  685. 

2.  Compare  spherical  pyramids  0-APB, 
0-BPC,  and  0-CPA  with  spherical  pyramids 
0-A'P'B',  O-B'P'C,  and  0-C'P'A\  respectively. 


SPHERICAL  PYRAMIDS  AND  WEDGES  453 

725.  Tlie  volume  of  a  triangular  spherical  pyramid  equals  one 
half  the  volume  of  a  spherical  wedge  whose  angle  is  the  spherical 
excess  of  the  base  of  the  pyramid. 

Tlie  proof  is  exactly  like  that  for  §  687. 

726.  If  the  radius  of  the  sphere  is  r  and  the  excess  of  the  base 
of  a  triangular  spherical  pyramid  is  Ey  and  the  volume  of  the 
spherical  pyramid   is  v,  then 

iTT^f^^^rr^.  §720 

2    270         540 

727.  The  same  formula  may  be  employed  to  find  the  volume 
of  any  spherical  pyramid,  with  the  understanding  that  E  is 
the  spherical  excess  of  the  base  of  the  pyramid,  measured  in 
degrees. 

728.  In  the  case  of  any  spherical  pyramid,  the  area  of  the 

base  is  — — "^  (§  689).     Hence  the  volume  of  any  spherical  pyra- 
loO 

mid  is  one  third  the  area  of  its  base  multiplied  by  the  radius 

of  the  sphere. 

Ex.  124.  Find  the  volume  of  a  triangular  spherical  pyramid  the 
angles  of  whose  base  are  92°,  119^,  and  134°,  if  the  volume  of  the  sphere 
is  192. 

Ex.  125.  Find  the  volume  of  a  quadrangular  spherical  pyramid,  the 
angles  of  whose  base  are  107°,  118°,  134°,  and  146°,  if  the  diameter  of  the 
sphere  is  12. 

Ex.  126.  The  volume  of  a  triangular  spherical  pyramid,  the  angles 
of  whose  base  are  105°,  126°,  and  147°,  is  60^.  What  is  the  volume  of  the 
sphere  ? 

Ex.  127.  Find  the  volume  of  a  pentagonal  spherical  pyramid  the 
angles  of  whose  base  are  109°,  128°,  137'',  163°,  and  168°,  if  the  volume  of 
the  sphere  is  180. 

Ex.  128.  The  volume  of  a  quadrangular  spherical  pyramid,  the 
angles  of  whose  base  are  110°,  122°,  136°,  and  146°  is  12 J.  What  is  the 
volume  of  the  sphere  ? 

Ex.  129.  What  is  the  angle  of  the  base  of  a  spherical  wedge  whose 
volume  is  Y"  """^  i^  ^^6  radius  of  the  sphere  is  4  ? 


454 


SOLID   GEOMETRY 


SUPPLEMENTARY  EXERCISES 

Ex.  1.  Two  planes  DEF  and  GEF  intersect  in 
line  EF.  A  is  any  point  in  plane  GEF.  If  ylC  be 
drawn  perpendicular  to  EF^  and  AB  perpendicular  to 
plane  DEF^  prove  the  plane  determined  hy  AC  and 
BC  perpendicular  to  EF. 


Ex.  2.  Prove  that  the  line  joining  the  mid-points  of  one  pair  of 
opposite  sides  of  a  quadrilateral  in  space  bisects  the  line  joining  the  mid- 
points of  the  other  pair  of  sides. 

Ex.  3.  If  two  intersecting  planes  pass  through  two  parallel  lines, 
their  intersection  is  parallel  to  the  parallel  lines. 

Ex.  4.  If  three  planes  intersect  in  pairs,  the  lines  of  intersection  are 
either  parallel  or  concurrent. 

Suggestions. — Case  (a)  1.  Assume  that  two  lines  of  intersection  meet  at 
a  point  P.  2.  Prove  that  P  is  on  the  third  line  of  intersection.  Case  (6) 
1.  Assume  that  two  lines  of  intersection  are  parallel.  2.  Prove  that  the  third 
line  of  intersection  is  parallel  to  the  other  two  by  an  indirect  proof. 

Ex.  5.     Prove  that  a  line  parallel  to  each  of  two  inter- 
secting planes  is  parallel  to  their  intersection. 

Hyp.  AB  is  II  planes  PR  and  QS. 

Con.  AB  II  QB. 

Suggestion.  —  Pass  a  plane  through  AB\\PR;    then  use    ^ 
§471. 

Ex.  6.  If  a  plane  be  drawn  through  a  diago- 
nal of  a  parallelogram  the  perpendiculars  to  it  from 
the  extremities  of  the  other  diagonal  are  equal. 

Hyp.  ABCD  is  a  O. 

BG  and  DH 1.  plane  AECF. 

Con.  BG  =  DH. 

Ex.  7.  D  is  any  point,  in  perpendicular  AF  from 
A  to  side  BC  ot  triangle  ABC.  If  line  I)E  be  drawn 
perpendicular  to  the  plane  of  ABC,  and  line  GH  be 
drawn  through  E  parallel  to  BC,  prove  line  AE  per- 
pendicular to  GH. 

Suggestion.  — FroYe  BC  1  to  plane  AED  and  then  GH 
1  plane  AED. 


SUPPLEMENTARY   EXERCISES 


455 


Ex.  8.  (a)  Through  a  line  which  is  parallel  to  a  plane,  a  plane  can 
be  drawn  parallel  to  the  given  plane. 

(6)  Is  the  construction  possible  if  the  given  line  is  not  parallel  to  the 
given  plane  ? 

Ex.  9.  If  a  plane,  parallel  to  the  edge  of  a  diedral  angle,  intersects 
ilie  faces  of  the  angle,  its  intersections  with  the  faces  are  parallel  to  the 
edge  and  to  each  other. 

Ex.  10.  If  a  straight  line  and  a  plane  are  both 
perpendicular  to  a  given  plane,  they  are  parallel,  un- 
less the  line  lies  in  the  plane. 

Hyp.     CB  ±  plane  PB  ;  plane  MN 1.  plane  PB. 
Con.  CBWMN. 


/ 1 

f- 

-,E 

/ 

B 

,  / 

^- 

y 

E 

Suggestions.  — \.  Draw  CD  1  RQ,  and  let  the  plane    

determined  by  CB  and  CD  meet  MN  in  DE.  R  ^ 

2.  Prove  CB  \\  DE  and  hence  ||  plane  MN  by  §  466. 

Ex.  11.  If  a  plane  is  perpendicular  to  one  of  two  perpendicular 
planes,  its  intersection  with  the  other  plane  is  also  perpendicular  to  the 
first  plane. 

Ex.  12.  From  any  point  E  within  diedral  angle 
CABD.,  EF  and  EG  are  drawn  perpendicular  to  faces 
ABC  and  ABD^  respectively,  and  G^IZ"  perpendicular  to 
face  ABC  at  //.     Prove  i^^is  perpendicular  to  AB. 

Suggestion  —  Prove  that  FII  lies  in  the  plane  of  EF 
and  EG,  by  §  498;  also  consider  the  relation  of  AB  and 
plane  GEF. 

Ex.  13.  If  7? C  is  the  projection  of  line  AB  upon 
plane  MN,  and  BD  and  BE  be  drawn  in  the  plane 
making  Z  CBD  =  Z  GBE,  prove  ZABD  =  ZABE. 

Suggestions.  — 1.  Lay  off  BD  =  BE,  and  draw  lines 
AD,  AE,  CD,  and  CE. 

2.  Prove  A  ABD  and  ABE  congruent. 

Ex.  14.  If  a  line  is  perpendicular  to  one  of  two  intersecting  planes, 
its  projection  upon  the  other  is  perpendicular  to  the  intersection  of  the 
two  planes. 

Suggestion.  — U  EG  1  plane  AD,  and  FII  is  the  projection  of  EG  on 
plane  BC,  prove  FH 1  AB. 

Ex.  15.  If  a  straight  line  intersects  two  parallel  planes,  it  makes 
equal  angles  with  them. 


456 


SOLID   GEOMETRY 


Ex.  16.  The  base  of  a  rectangle  ABCT)  is  10, 
and  its  altitude  8.  Side  10  is  parallel  to  plane  QB. 
Side  8  makes  an  angle  of  60°  with  QB.  Find  the 
area  of  the  projection  of  O  ABCD  on  plane  QR, 
correct  to  three  decimal  places. 

Ex.  17.  An  equilateral  A  ABC,  whose  area  is  25, 
has  its  side  BC  parallel  to  a  plane  QB.  The  plane 
of  A  ABC  makes  an  angle  of  45°  with  plane  QB. 
Find  the  area  of  the  projection  of  A  ABC  on 
plane  BQ. 

Ex.  18.     Find  the  lateral  area  of  a  regular  triangular  prism  each  side 
of  whose  base  is  5  and  whose  altitude  is  8. 

Ex.  19.     Prove  that  the  upper  base  of  a  truncated  parallelepiped  is  a 
parallelogram. 

Ex.  20.  Prove  that  the  sum  of  two  opposite 
lateral  edges  of  a  truncated  parallelopiped  is  equal 
to  the  sum  of  the  other  two  lateral  edges. 

Suggestions.  —  1.  What  kind  of  figure  is 
AA'C'C? 

2.  Compare  AA'  +  CC  with  00'. 

Ex.  21.  Prove  that  the  perpendicular  drawn 
to  the  lower  base  of  a  truncated  right  triangular 
prism  from  the  intersection  of  the  medians  of  the 
upper  base,  is  equal  to  one  third  the  sum  of  the 
lateral  edges. 

Suggestion.  —  Let  P  be  the  mid-point  of  DL,  and 
draw  PQ  1  ABC',  express  LM  in  terms  oi  FQ  and 
GN. 

Ex.  22.  Prove  that  the  sum  of  the  squares  of  the 
four  diagonals  of  a  parallelopiped  is  equal  to  the  sum 
of  the  squares  of  its  twelve  edges. 

Suggestion.  — 1^C2A\  Ex.  142,  Book  III,  p.  184. 

Ex.  23.     Determine  the  approximate  area  of  the  base  of  a  bin  6  ft. 
deep  that  will  hold  250  bu.  of  grain.     (One  bu.  =  2150.42  cu.  in.) 

Ex.  24.     Find  the  edge  of  a  cube  equivalent  to  a  rectangular  parallelo- 
piped whose  dimensions  are  9  in.,  1  ft.  9  in.,  and  4  ft.  1  in. 


SUPPLEMENTARY   EXERCISES  457 

Ex.  25.  Find  the  volume  of  a  rectangular  parallelopiped,  the  dimen- 
sions of  whose  base  are  14  and  9,  and  the  area  of  whose  entire  surface  is 
620. 

Ex.  26.  The  diagonal  of  a  cube  is  8  VS.  Find  its  volume,  and  the 
area  of  its  entire  surface. 

Suggestion.  —  Represent  the  length  of  the  edge  by  x. 

Ex.  27.     Find  the  dimensions  of  the  base  of  a  rectangular  parallelo- 
piped, the  area  of  whose  entire  surface  is  320,  volume  336,  and  altitude  4. 
Suggestion.  — Represent  the  dimensions  of  the  base  by  x  and  y. 

Ex.  28.  Find  the  area  of  the  entire  surface  of  a  rectangular  parallelo- 
piped, the  dimensions  of  whose  base  are  11  and  13,  and  volume  858. 

Ex.  29.  A  trench  is  124  ft.  long,  2\  ft.  deep,  6  ft.  wide  at  the  top,  and 
5  ft.  wide  at  the  bottom.     How  many  cubic  feet  of  water  will  it  contain  ? 

Ex.  30.  Prove  that  the  volume  of  any  oblique  prism  is  equal  to  the 
product  of  the  area  of  a  right  section  by  the  length  of  a  lateral  edge. 

Ex.  31.  Prove  that  the  volume  of  a  regular  prism  is  equal  to  its 
lateral  area  multiplied  by  one  half  the  apothem  of  the  base. 

Ex.  32.  The  volume  of  a  right  prism  is  2310,  and  its  base  is  a  right 
triangle  whose  legs  are  20  and  21,  respectively.     Find  its  lateral  area. 

Ex.  33.     The  lateral  area  and  volume  of  a  regular  hexagonal  prism  are 
60  and  15  VS,  respectively.     Find  its  altitude,  and  one  side  of  its  base. 
Suggestion.  —  Represent  the  altitude  by  x,  and  the  side  of  the  base  by  y. 

Ex.  34.  The  altitude  of  a  pyramid  is  20  in.,  and  its  base  is  a  rectangle 
whose  dimensions  are  10  in.  and  15  in. ,  respectively.  What  is  the  distance 
from  the  vertex  of  a  section  parallel  to  the  base,  whose  area  is  54  sq.  in.? 

Ex.  35.  At  what  distance  from  the  altitude  must  a  plane  parallel  to 
the  base  be  drawn  so  that  the  area  of  the  section  will  be  one  half  the  base  ? 

Ex.  36.  In  Ex.  35,  replace  the  fraction  ^  by  the  fraction  ^  and  solve 
the  resulting  exercise. 

Ex.  37.  In  Ex.  35,  replace  the  fraction  ^  by  the  fraction  ^  and  solve 
the  resulting  exercise. 

Ex.  38.  Prove  that  the  volum/B  of  a  regular  pyramid  is  equal  to  its 
lateral  area,  multiplied  by  one  third  the  distance  from  the  center  of  its 
base  to  any  lateral  face. 

Suggestion.  —  Pass  planes  through  the  lateral  edges  and  the  center  of  the 


458 


SOLID   GEOMETRY 


Ex.  39.    Find  the  lateral  edge,  lateral  area,  and  volume  of  a  frustum 
of  a  regular  quadrangular  pyramid,  the  sides  of 
whose  bases  are  17  and  7,  respectively,  and  whose 
altitude  is  12. 

Suggestion. — Let  ABB' A'  be  a  lateral  face  of 
the  frustum,  and  0  and  0'  the  centers  of  the  bases; 
draw  lines  OC 1.  AB,  O'C  1  A'B',  CD  1  OC,  and 
A'E  1  AB\  also  lines  00'  and  CC. 

Ex.  40.  The  bases  of  a  frustum  of  a  pyramid  are  rectangles,  whose 
sides  are  27  and  15,  and  9  and  5,  respectively,  and  the  line  joining  their 
centers  is  perpendicular  to  each  base.  If  the  altitude  of  the  frustum  is 
12,  find  its  lateral  area  and  volume. 

Ex.  41.  Find  the  lateral  area  and  volume  of  a  frustum  of  a  regular 
triangular  pyramid,  the  sides  of  whose  bases  are  12  and  6,  respectively, 
and  whose  lateral  edge  is  6. 

Ex.  42.  The  altitude  and  lateral  edge  of  a  frustum  of  a  regular 
triangular  pyramid  are  8  and  10,  respectively,  and  each  side  of  its  upper 
base  is  2\/3.     Find  its  volume  and  lateral  area. 

Ex.  43.  Find  the  volume  of  the  rectangular  prismoid  the  sides  of 
whose  bases  are  10  and  7,  and  6  and  5,  respectively,  and  whose  altitude 
is  9. 

Ex.  44.  The  volume  of  a  triangular  prism  is  equal  to  a  lateral  face, 
multiplied  by  one  half  its  perpendicular  distance  from  any  point  in  the 
opposite  lateral  edge. 

Suggestion.  — Draw  a  rt.  section  of  the  prism,  and  apply  §  577. 

Ex.  45.     Prove  that  the  volume  of    a  truncated  parallelopiped   is 
equal  to  the  area  of  a  right  section  multiplied  by 
one  fourth  the  sum  of  the  lateral  edges. 

Ex.  46.  Prove  that  a  plane  passed  through 
the  center  of  a  parallelopiped  divides  it  into  two 
equal  solids. 

Ex.  47.  The  volume  of  a  truncated  parallelo- 
piped is  equal  to  the  area  of  a  right  section,  mul- 
tiplied by  the  distance  between  the  centers  of  the 


Suggestion.  —  By  Ex.  45,  the  distance  between  the  centers  of  the  bases 
may  be  proved  equal  to  one  fourth  the  sum  of  the  lateral  edges. 


SUPPLEMENTARY   EXERCISES  459 

Ex.  48.  How  many  square  feet  of  heating  surface  are  there  in  a  hot- 
water  conducting  pipe  9  feet  long  and  2  inches  in  outside  diameter  ? 

Ex.  49.     Determine  the  lateral  area  of  the   right  circular  cylinder 
formed  by  revolving  a  rectangle,  having  base  b  and  altitude  h, 
(a)  about  its  base  ;  (6)  about  its  altitude. 

Ex.  50.  The  lateral  area  of  a  cylinder  of  revolution  is  120  tt.  The 
area  of  the  base  is  36  tt.     Find  the  altitude. 

Ex.  51.  The  cross  section  of  a  tunnel,  2^  mi.  in  length,  is  in  the  form 
of  a  rectangle  0  yd.  wide  and  4  yd.  high,  surmounted  by  a  semicircle 
whose  diameter  is  equal  to  the  width  of  the  rectangle  ;  how  many  cubic 
yards  of  material  were  taken  out  in  its  constniction  ?     (tt  =  3.1416.) 

Ex.  52.  What  must  be  the  length  in  inches  of  a  10-gal.  gasoline  tank 
which  is  10  in.  in  diameter  ? 

Ex.  53.  Determine  the  volume  generated  when  a  rectangle  of  base  b 
and  altitude  h 

(a)  revolves  about  its  side  b  ;    (6)  revolves  about  its  side  h. 

Ex.  54.  Two  right  circular  cylinders  have  equal  altitudes,  but  the 
radius  of  the  base  of  the  one  is  double  the  radius  of  the  base  of  the  other. 
Compare  (a)  their  lateral  areas  ;  (6)  their  volumes. 

Ex.  55.  A  regular  hexagonal  prism  is  inscribed  in  a  right  circular 
cylinder  whose  altitude  is  10  in.  and  the  radius  of  whose  base  is  3  in. 
Determine  the  difference  between  the  volumes  of  the  prism  and  cylinder. 

Ex.  56.  Prove  that  the  volume  of  a  cylinder  of  revolution  is  equal  to 
its  lateral  area  multiplied  by  one  half  the  radius  of  its  base. 

Ex.  57. .  Express  by  a  formula  the  volume  of  a  round  cast-iron  column 
of  length  I  ft.,  thickness  t  in.,  and  outside  diameter  d  in. 

Ex.  58.  Given  the  radius  of  the  base  B  and  the  total  area  T  of  a 
cylinder  of  revolution,  find  its  volume. 

(Find  S'from  the  equation  T=2  irEH  +  2  -n-R^.) 

Ex.  59.  Given  the  diameter  of  the  base  D  and  the  volume  Fof  a 
cylinder  of  revolution,  find  its  lateral  area  and  total  area. 

Ex.  60.  The  volume  of  a  circular  cone  is  F.  What  is  the  effect 
upon  the  volume  : 

(a)  if  the  radius  of  the  base  is  doubled  ? 

(6)  if  the  altitude  is  doubled  ? 

(c)  if  both  the  radius  and  the  altitude  are  doubled  ? 

Ex.  61.  The  altitude  of  a  cone  of  revolution  is  27  in.,  and  the  radius 
of  its  base  is  16  in.  What  is  the  diameter  of  the  base  of  an  equal  cylin- 
der, whose  altitude  is  16  in.  ? 


460  SOLID   GEOMETRY 

Ex.  62.  A  plane  is  passed  parallel  to  the  base  of  a  circular  cone  so 
as  to  bisect  the  altitude.  What  is  the  ratio  of  the  two  parts  into  which 
the  given  cone  is  divided  ? 

Ex.  63.  Determine  the  lateral  area  of  a  right  circular  cone  whose 
volume  is  320  tt  cu.  in.,  and  whose  altitude  is  15  in. 

Ex.  64.  Determine  the  volume  of  a  cone  of  revolution  whose  slant 
height  is  29  in.,  and  whose  lateral  area  is  580  tt  sq.  in. 

Ex.  65.  If  the  altitude  of  a  cone  of  revolution  is  three  fourths  the 
radius  of  its  base,  the  volume  is  equal  to  its  lateral  area  multiplied  by  one 
fifth  the  radius  of  its  base. 

Ex.  66.  Given  the  altitude  H  and  the  volume  T  of  a  right  circular 
cone.     Derive  the  foi-mula  for  the  lateral  area  in  terms  of  V  and  H, 

Ex.  67.  Given  the  slant  height  L  and  the  lateral  area  S  of  a  right 
circular  cone.     Derive  the  formula  for  its  volume  in  terms  of  S  and  L. 

Ex,  68.  Find  the  lateral  area  of  the  frustum  of  a  right  circular  cone, 
whose  altitude  is  8  in.,  if  the  radii  of  its  bases  are  6  in.  and  3  in.,  respec- 
tively. 

Ex.  69.  A  tapering  hollow  iron  column,  1  in.  thick,  is  24  ft.  long,  10 
in.  in  outside  diameter  at  one  end  and  8  in.  in  diameter  at  the  other. 
How  many  cubic  inches  of  metal  are  there  in  it  ? 

Ex.  70.  Prove  that  a  frustum  of  a  circular  cone  is  equal  to  three 
cones  whose  common  altitude  is  the  altitude  of  the  frustum,  and  whose 
bases  equal  the  lower  base,  the  upper  base,  and  the  mean  proportional 
between  the  bases  of  the  frustum. 

Ex.  71.  The  area  of  the  entire  surface  of  a  frustum  of  ,a  cone  of 
revolution  is  306  ir  sq.  in.,  and  the  radii  of  its  bases  are  11  in.  and  5  in., 
respectively.     Find  the  lateral  area  and  the  volume  of  it. 

Ex.  72.  The  volume  of  a  frustum  of  a  right  circular  cone  is  6020  tt 
cu.  in.,  its  altitude  is  60  in.,  and  the  radius  of  its  lower  base  is  15  in. 
Find  the  radius  of  the  upper  base  and  its  lateral  area. 

Ex.  73.  Find  the  diameter  and  the  area  of  the  surface  of  a  sphere 
whose  volume  is  ^^^-^ir  cu.  in. 

Ex.  74.  The  altitude  of  a  frustum  of  a  cone  of  revolution  is  3^,  and 
the  radii  of  its  bases  are  5  and  3  ;  what  is  the  diameter  of  an  equal 
sphere  ? 

Ex.  75.  Find  the  area  of  the  surface  and  the  volume  of  a  sphere  cir- 
cumscribing a  cylinder  of  revolution,  the  radius  of  whose  base  is  9,  and 
whose  altitude  is  24. 


SUPPLEMENTARY  EXERCISES 


461 


Ex.  76.  A  cone  of  revolution  is  inscribed  in  a  sphere  whose  diameter 
is  ^  the  altitude  of  the  cone.  Prove  that  its  lateral  surface  and  volume 
are,  respectively,  f  and  /^  the  surface  and  volume  of  the  sphere. 

Ex.  77.  Given  the  area  of  the  surface  of  a  sphere  S  to  find  its 
volume. 

Ex.  78.     Given  the  volume  of  a  sphere  V  to  find  the  area  of  its  surface. 

Ex.  79.  A  portion  of  a  plane  bounded  by  an  eqiiilateral  triangle, 
whose  side  is  6,  revolves  about  one  of  its  sides  as  an  axis.  Find  the  area 
of  the  entire  surface,  and  the  volume  of  the  solid  generated. 

Ex.  80.  A  circular  sector  whose  central  angle  is  45°  and  radius  12 
revolves  about  a  diameter  perpendicular  to  one  of  its  bounding  radii. 
Find  the  volume  of  the  spherical  sector  generated. 

Ex.  81.  A  portion  of  a  plane  bounded  by  a  right  triangle,  whose 
legs  are  a  and  6,  revolves  about  its  hypotenuse  as  an  axis.  Find  the  area 
of  the  entire  surface,  and  the  volume  of  the  solid  generated. 

Ex.  82.  A  portion  of  a  plane  bounded  by  an  equilateral  triangle, 
whose  altitude  is  h,  revolves  about  one  of  its  altitudes  as  an  axis.  Find 
the  area  of  the  surface,  and  the  volume  of  the  solid  generated. 

Ex.  83.     A  portion  of  a  plane  bounded  by  an  equilateral  triangle, 
whose   side  is  a,    revolves    about   a   straight  line  drawn 
through  one  of  its  vertices  parallel  to  the  opposite  side. 
Find  the  area  of  the  entire  surface,  and  the  volume  of  the 
solid  generated. 

(The  solid  generated  is  the  difference  of  the  cylinder 
generated  by  BCUG,  and  the  cones  generated  by  ABG 
and  ACH.) 

Ex.  84.  If  a  portion  of  a  plane  bounded  by  any  triangle  be  revolved 
about  an  axis  in  its  plane,  not  parallel  to  its  base,  which  passes  through 
its  vertex  without  intersecting  its  surface,  the  volume  of  the  solid  gen- 
erated is  equal  to  the  area  of  the  surface  generated  by  the  base,  multiplied 
by  one  third  the  altitude. 

Ex.  85.  If  a  portion  of  a  plane  bounded  by  any  triangle  be  revolved 
about  an  axis  which  passes  through  its 
vertex  parallel  to  its  base,  the  volume  of 
the  solid  generated  is  equal  to  the  area  of 
the  surface  generated  by  the  base,  multi- 
plied by  one  third  the  altitude. 

Ex.  86.  Find  the  volume  of  a  spher- 
ical sector,  the  altitude  of  whose  base  is 
12,  the  diameter  of  the  sphere  being  25. 


IMPORTANT  DEFINITIONS  AND  THEOREMS   OF  PLANE 
GEOMETRY 

§  8.    {a)  One  and  only  one  straight  line  can  be  drawn  through  two 
points. 

(h)  A  straight  line  can  be  extended  indefinitely  in  each  direction. 

§  11,    Two  straight  lines  can  intersect  at  only  one  point. 

§  14.    The  straight  line  segment  is  the  shortest  line  between  two  points. 

§  16.    A  circle  is  a  closed  curved  line  (in  a  plane)  all  points  of  which 
are  equidistant  from  a  point  within  called  the  center. 

§  17.    All  radii  of  the  same  circle  or  of  equal  circles  are  equal. 

§  20.    An  angle  is  the  figure  formed  by  two  rays  drawn  from  the  same 
point. 

§  24.   Adjacent  angles  are  two  angles  that  have  a  common  vertex  and 
a  common  side  between  them. 

§  26.    If  one  straight  line  meets  another  straight  line  so  that  the  adjacent 
angles  formed  are  equal,  each  of  these  angles  is  a  right  angle. 

§  27.    All  right  angles  are  equal. 

§  29.    Two  lines  are  perpendicular  if  they  form  a  right  angle. 

§  34.   The  sum  of  all  the  successive  adjacent  angles  around  a  point  on 
one  side  of  a  straight  line  is  one  straight  angle. 

§  35.   The  sum  of  all  the  successive  adjacent  angles  around  a  point  is 
two  straight  angles. 

§  36.    Two  angles  are  complementary  if  their  sum  equals  a  right  angle. 

§  37.    Complements  of  the  same  angle  or  of  equal  angles  are  equal. 

§  38.    Two  angles  are  supplementary  if  their  sum  is  equal  to  a  straight 
angle. 

§  39.    If  two  adjacent  angles  have  their  exterior  sides  in  a  straight  line, 
they  are  supplementary. 

§  40.   If  two  adjacent  angles  are  supplementary,  their  exterior  sides 
are  in  a  straight  line. 

§  41.   Supplements  of  the  same  angle  or  of  equal  angles  are  equal. 

462 


i 


INDEX 


Altitude,  of  a  O,  §  131 ;  of  a  trape- 
zoid, §  145  ;  of  a  A,  §  85. 

Analysis,  §  236,  §  347. 

Angle,  §  20;  acute,  §  30;  central, 
§  180  ;  diedral,  §  484 ;  inscribed, 
§  216  ;  intercepted,  §  180 ;  obtuse, 
§  30  ;  polyedral,  §  507  ;  right,  §  26; 
spherical,  §  647  ;  straight,  §  31  ; 
triedral,  §  508 ;  tetraedral,  §  508 ; 
between  two  intersecting  curves, 
§  047  ;  between  a  line  and  a  plane, 
§  506  ;  bisector  of,  §  23  ;  cosine  of, 
§  301  ;  measure  of,  §  28;  sides  of, 
§  20  ;  sine  of,  §  300  ;  tangent  of, 
§  301 ;  vertex  of,  §  20  ;  bisect  an, 
§  74  ;  construct  an,  §  75 ;  meas- 
ure an,  by  protractor,  §  44  ;  ex- 
terior, of  a  A,  §  86  ;  of  elevation, 
§  303  ;  of  depression,  §  S03. 

Angles,  adj.,  §  24;  alt. -int.,  §  92; 
alt.-ext. ,  §  92  ;  complementary, 
§  30  ;  corresponding,  §  92  ;  equal, 
§  21  ;  exterior  of  lis,  §  92  ;  homolo- 
gous of  /S^,  §  65  ;  interior  of  lis, 
§  92  ;  supplementary,  §  38 ;  supp.- 
adj.,  §  39  ;  vertical,  §  42. 

Antecedent,  §  242. 

Apothem,  §  360. 

Arc,  §  179 ;  minor,  §  179  ;  major, 
§  179  ;  measure  of,  §  214  ;  inter- 
cepted, §  180 ;  subtended,  §  184. 

Area,  §  320 ;  of  a  O,  §  392,  §  416, 
§  417  ;  of  a  O,  §  331  ;  of  a  Q, 
§  330  ;  of  a  sector,  §  397  ;  of  a 


segment  of  a  O,  §  398  ;  of  a  spher- 
ical surface,  §  672  ;  of  a  trapezoid, 
§  337,  §  338 ;  of  a  A,  §  333,  §  335  ; 
lateral,  of  a  prism,  §  531  ;  lateral, 
of  a  regular  pyramid,  §  566  ;  lat- 
eral, of  a  rt.  circ.  cylinder,  §  599 ; 
lateral,  of  a  rt.  circ.  cone,  §  613. 

Axiom,  §  51  ;  of  congruence,  §  61 ; 
of  limits,  §  402  ;  of  lis,  §  90. 

Axioms,  list  of,  §  51,  §  61,  §  90, 
§158. 

Axis,  of  a  circle  of  a  sphere,  §  634 ; 
of  a  circ.  cone,  §  605  ;  of  a  circ. 
cylinder,  §  593;  of  symmetry, 
§  427. 

Base,  of  a  O,  §  131 ;  of  a  A,  §  68  ; 
of  an  isosceles  A,  §  68. 

Bases,  of  a  O,  §  131  ;  of  a  trape- 
zoid, §  145. 

Bisect  a  segment,  §  78  ;  an  angle, 
§74. 

Broken  line,  §  5. 

Center,  of  a  O,  §  16;  of  gravity, 
§  172  ;  of  a  O,  §  137  ;  of  a  regu- 
lar polygon,  §  360 ;  of  symmetry, 
§  426. 

Chord,  §  184. 

Circle,  §  16 ;  arc  of,  §  179 ;  area  of, 
§  392,  §  416,  §  417  ;  center  of, 
§  16  ;  chord  of,  §  16 ;  circumfer- 
ence of,  §  389  ;  diameter  of,  §  16  ; 
interior  of,  §  174 ;  radius  of,  §  16  ; 


463 


464 


INDEX 


sector  of,  §  396 ;  segment  of, 
§  398  ;  circumscribed,  §  178  ; 
great,  of  a  sphere,  §  634 ;  in- 
scribed,  §  225 ;  small,  of  a  sphere, 
§  634. 

Circles,  concentric,  §  176 ;  equal, 
§  17  ;  tangent,  §  204 ;  tangent 
internally,  §  204;  tangent  ex- 
ternally, §  204. 

Circumcenter,  §  170. 

Circumference,  §  389. 

Circumscribed,  circle,  §  178;  poly- 
gon, §  225. 

Commensurable,  §  211. 

Complement,  §  36. 

Conclusion,  §  52. 

Concurrent  lines,  §  168. 

Cone,  §  604  ;  altitude  of  a,  §  604  ; 
axis  of  a  circular,  §  605 ;  base  of 
a,  §  604 ;  circular,  §  605  ;  frustum 
of  a,  §  605;  lateral  surface  of  a, 
§  604;  rt.  circ,  §  605. 

Cones  of  revolutipn,  §  625. 

Congruence,  §  59 ;  axiom  of,  §  61. 

Conical  surface,  §  603  ;  directrix  of 
a,  §  603  ;  element  of  a,  §  603 ; 
generatrix  of  a,  §  603  ;  vertex  of 
a,  §  604. 

Consequent,  §  242. 

Constant,  §  401,  §  542. 

Converse,  §  104. 

Co-planar,  §  444. 

Corollary,  §  71. 

Cosine,  §  301. 

Cylinder,  §  588  ;  altitude  of  a,  §  588  ; 
bases  of  a,  §  588  ;  circular,  §  589  ; 
elements  of  a,  §  588  ;  lateral  area 
of  a,  §  597  ;  lateral  area  of  a  rt. 
circ,  §  599  ;  lateral  surface  of  a, 
§  588  ;  total  surface  of  a,  §  588 ; 
right,  §  589  ;  volume  of  a,  §  597, 
§601. 


Degree,  of  angle,  §  28  ;  of  arc,  §  214. 
Diedral  angle,  §  484 ;   edge  of  a, 

§  484  ;  faces  of  a,  §  484  ;  plane  A 

of  a,  §  486.  I 

Diedral  angles,  adj.,  §  485;  equal,    | 

§  489  ;  vertical,  §  485.  ^ 

Dimensions  of  a  rect.  parallelopiped, 

§541. 
Directrix,  of  a  conical  surface,  §  603 ; 

of  a  cylindrical  surface,  §  586. 
Distance,  between  II  planes,  §  478 ; 

between  points  on  the  surface  of 

a  sphere,  §  641  ;  from  a  point  to 

a  line,  §  84  ;  polar,  §  644. 
Dodecaedron,  §  527. 

Edge,  of  a  half  plane,  §  483. 
Equal  angles,  §  21 ;  ©,  §  17  ;  seg- 
ment, §  13  ;  surfaces,  §  322. 
Exterior  angle,  §  86. 
Extreme  and  mean  ratio,  §  377. 
Extremes,  §  244. 

Foot  of  a  line  on  a  plane,  §  449. 

Frustum,  of  a  pyramid,  §  560;  of  a 
cone,  §604 ;  of  a  pyramid  in- 
scribed in  a  frustum  of  a  cone, 
§611. 

Generatrix,  of  a  conical  surface, 
§  603  ;  of  a  cylindrical  surface, 
§  586. 

Great  circle,  §  634. 

Half-plane,  §  483  ;  edge  of  a,  §  483. 

Hexaedron,  §  527. 

Homologous   parts,    §  65 ;    sides  of 

similar  triangles,  §  278. 
Hypotenuse,  §  107. 
Hypothesis,  §  52. 

Icosaedron,  §  527. 
In-center,  §  169,  §  226. 


INDEX 


465 


Inclination  of  a  line  to  a  plane,  §  50(). 
Incommensurable,      §  211  ;     cases, 

§  423,  §  424,  §  425. 
Indirect  method,  §  94. 
Inscribed,    angle,    §  216 ;    polygon, 

§178. 
Interior  of  closed  line,  §  6. 
Intersection,  of  lines,  §  10  ;  of  two 

surfaces,  §  448. 
Isoperimetric,  §  432. 

Lateral  area,  of  a  frustum  of  a  reg. 
pyramid,  §  667  ;  of  a  frustum  of  a 
rt.  circ.  cone,  §617;  of  a  cylinder, 
§  698 ;  of  a  prism,  §  631 ;  of  a 
reg.  pyramid,  §  566  ;  of  a  rt.  circ. 
cone,  §  613. 

Legs  of  a  rt.  triangle,  §  107. 

Length  of  a  circle,  §  388,  §  407, 
§412. 

Limit,  §  401. 

Line,  broken,  §  5  ;  curved,  §  5  ; 
II  to  a  plane,  §  464  ;  ±  to  a  plane, 
§  452  ;  straight,  §  6 ;  of  centers, 
§  206. 

Lines,  concurrent,  §  168. 

Loci,  method  of  attacking,  §  238 ; 
intersection  of,  §  340  ;  construc- 
tion by,  §  241. 

Locus  of  points,  §  229,  §  518. 

Lune,  §  675  ;  angle  of  a,  §  675. 

Maximum,  §  433. 

Means,  §  244. 

Measure,  numerical,  §210  ;  common, 

§  211 ;  of  arc,   §  214  ;  of  central 

angle,  §216;  of  inscribed  angle, 

§217. 
Median,  of  A,  §  79 ;  of  trapezoid, 

§  145. 
Mid-point  of  a  segment,  §  16. 
Minimum,  §  433. 


Octaedron,  §  527. 
Ortho-center,  §  171. 

Parallel  lines,  §  89  ;  axiom  of,  §  90  ; 
construction  of,  §  99. 

Parallelogram,  §  131 ;  altitude  of  a, 
§  131  ;  bases  of  a,  §  131. 

Parallelopiped,  §  537 ;  rt.,  §  537  ; 
rect.,  §  537  ;  center  of  a,  Ex.  15, 
p.  353. 

Perpendicular,  §  29  ;  construction 
of,  §  80,  §  82. 

Perpendicular-bisector,  §  76  ;  con- 
struction of,  §  78. 

Plan  of  a  proof,  §  117. 

Plane,  §  443  ;  determined,  §  444  ; 
Z  of  a  diedral  Z,  §  486  ;  ±  to  a 
line,  §  452  ;  tangent  to  a  cylinder 
and  to  a  cone,  §  622. 

Planes,  parallel,  §  465  ;  perpendicu- 
lar, §  494. 

Point,  §  4,  note  p.  27  ;  mid-,  §  15 ; 
of  contact,  §  197  ;  of  tangency, 
§  197  ;  projection  of  a,  §  308, 

Points,  of  intersection,  §  10 ;  locus 
of,  §  228,  §  229. 

Polar  distance,  §  644. 

Poles  of  a  O  of  a  sphere,  §  634. 

Polyedra,  similar,  §  582. 

Polyedral  angle,  §  507  ;  diedral  A 
of  a,  §  507  ;  edges  of  a,  §  507 ; 
faces  of  a,  §  607 ;  face  A  of  a, 
§  507;  vertex  of  a,  §  507. 

Polyedral  angles,  congruent,  §  610, 
symmetrical,  §  511 ;  vertical,  §  509. 

Polyedron,  §  626 ;  diagonal  of  a, 
§  526  ;  edges  of  a,  §  526  ;  faces  of 
a,  §  526  ;  regular,  §  679  ;  vertices 
of  a,  §  526. 

Polygon,  §  126  ;  .4  of,  §  126  ;  diag- 
onal of,  §  125 ;  interior  of,  §  125  ; 
perimeter    of,    §  126 ;    sides    of, 


466 


INDEX 


§  125  ;  vertices  of,  §  125  ;  circum- 
scribed, §  225  ;  concave,  §  126  ; 
convex,  §  126 ;  equiangular, 
§  128  ;  equilateral,  §  128  ;  in- 
scribed, §  178 ;  mutually  equi- 
angular, or  equilateral,  §  129 ; 
regular,  §  356 ;  similar,  §  272. 

Postulate,  §  56. 

Postulates,  list  of,  §  56,  §  62. 

Prism,  §  528  ;  altitude  of  a,  §  528 ; 
bases  of  a,  §  528 ;  lateral  area 
of  a,  §  528  ;  lateral  edges  of  a, 
§  528  ;  lateral  faces  of  a,  §  528  ; 
oblique,  §  530  ;  regular,  §  530 ; 
right,  §  530 ;  truncated,  §  630. 

Prismatoid,  §  573 ;  altitude  of  a, 
§  573 ;  bases  of  a,  §  573 ;  mid- 
section of  a,  §  573. 

Problem,  §  55. 

Projection,  of  a  point  on  a  line, 
§  308 ;  of  a  point  on  a  plane, 
§  503  ;  of  a  segment  on  a  line, 
§  309 ;  of  a  line  on  a  plane,  §  503. 

Proportion,  §  244 ;  by  alternation, 
§  253 ;  by  inversion,  §  254  ;  by 
composition,  §  255 ;  by  composi- 
tion and  division,  §  257  ;  by  divi- 
sion, §  256. 

Proportional,  fourth,  §  246  ;  third, 
§  247  ;  mean,  §  248. 

Proposition,  §  57. 

Protractor,  §  43 ;  use  of,  §  44, 
§46. 

Pyramid,  §  58  ;  altitude  of  a,  §  658 ; 
base  of  a,  §  568  ;  lateral  area  of 
a,  §  558  ;  lateral  edges  of  a,  §  558  ; 
lateral  faces  of  a,  §  568  ;  vertex 
of  a,  §  658  ;  frustum  of  a,  §  560 ; 
regular,  §  559 ;  truncated,  §  559  ,• 
inscribed  in  a  cone,  §  610. 

Quadrant,  of  circle,  §  176.  ] 


Radius,  of  a  O,  §  16  ;  of  circum- 
scribed O  of  a  A,  §  177  ;  of  in- 
scribed O  of  a  A,  §  226 ;  of  a 
regular  polygon,  §  360. 

Ratio,  §  212,  §  242  ;  of  similitude, 
§  272  ;  extreme  and  mean,  §  377. 

Ray,  §  19. 

Rectangle,  §  141. 

Regular  polyedron,  §  579 ;  pyramid, 
§559. 

Regular  polygon,  §  356 ;  apothem 
of,  §  360  ;  central  A  of,  §  360;  ra- 
dius of,  §  360  ;  construction  of, 
§  370  ;  vertex  Z  of,  §  360. 

Rhombus,  §  142. 

Scales,  §  299. 

Secant,  §  194 ;  whole,  §  285. 

Segment  of  a  circle,  §  398  ;  of  a  line, 
§  12  ;  external,  §  286  ;  internal, 
§386;  spherical,  §716;  projec- 
tion of  a,  §  309 ;  divided  exter- 
nally, §  306 ;  divided  harmoni- 
cally, §  307  ;  divided  internally, 
§269. 

Segments,  equal,  §  13 ;  divided  pro- 
portionally, §  260. 

Semicircle,  §  176. 

Sequence  of  polygons,  §  404. 

Sine  of  an  angle,  §  300. 

Slant  height,  of  a  frustum  of  a  reg. 
pyramid,  §  562  ;  of  a  frustum  of 
a  rt.  circ.  cone,  §  606 ;  of  a  reg. 
pyramid,  §  562 ;  of  a  rt.  circ. 
cone,  §  606. 

Solid,  §  529 ;  volume  of  a,  §  533. 

Solids,  equal,  §  534. 

Sphere,  §  629  ;  radius  of  a,  §  630 ; 
diameter  of  a,  §  630. 

Spheres,  tangent,  §  665. 

Spherical,  angle,  §  647 ;  excess, 
§  660;    pyramid,    §   722;    sector. 


INDEX 


467 


§  «90  ;  segment,  §  716  ;  surface, 
§  027 ;  triangle,  §  662  ;  wedge, 
§  718. 

Spherical  polygon,  §  651  ;  angle  of 
a,  §  051  ;  diagonal  of  a,  §  651  ; 
sides  of  a,  §  651. 

Spherical  polygons,  symmetrical, 
§681. 

Square,  §  143. 

Subtend,  §  184. 

Superposition,  §  62  ;  postulate  of, 
§62. 

Supplement,  §  38. 

Surface,  §  442  ;  closed,  §  523 ;  coni- 
cal, §  603  ;  convex,  §  524 ;  curved, 
§  522  ;  cylindrical,  §  586 ;  gen- 
erating a,  §  585  ;  spherical,  §  628. 

Symmetrical  spherical  polygons, 
§681. 

Symmetry  with  respect  to,  a  straight 
line,  §  429  ;  a  center,  §  428. 

Tangent,   §  197 ;   common,   §  203 ; 

external,  §  203;  internal,  §  203; 

of  angle,  §  301. 
Tangent  to  a  sphere,  a  plane,  §  662. 
Tetraedral  angle,  §  508. 
Tetraedron,  §  527. 
Theorem,  §  52 ;    proved   formally, 

§  54 ;  proved  informally,  §  53. 
Transversal,  §  92. 
Trapezoid,  §  145  ;  altitude  of,  §  145 ; 

area  of,  §  337  ;  bases  of,  §  146 ; 


isosceles,  §  146;  median  of, 
§146. 

Triangle,  §  47  ;  altitude  of,  §  85 ; 
base  of,  §  68;  median  of,  §  79; 
parts  of,  §  47  ;  sides  of,  §  47  ;  ver- 
tical Z.  of,  §  68  ;  vertices  of,  §  47 ; 
bi-rectangular,  §  661 ;  circum- 
scribed, §  225  ;  equiangular,  §  68  ; 
equilateral,  §  68 ;  inscribed,  §  178 ; 
isosceles,  §  68 ;  isosceles  right, 
§  107  ;  right,  §  107  ;  scalene,  §  68  ; 
tri-reetangular,  §  661 ;  construc- 
tion of,  §  232. 

Triedral  angle,  §  508. 

Trigonometric  tables,  §  302. 

Trigonometry,  §  300. 


Unit,  of  surface,  §  321 
§  28  ;  of  arc,  §  214. 


of  angle, 


Variable,  §  401,  §  542  ;  limit  of  a, 
§542. 

Vertical  angle,  of  a  A,  §  68 ;  of  an 
isosceles  A,  §  68. 

Volume  of  a  cone,  §  612 ;  of  a  rt. 
circ.  cone,  §  615  ;  of  a  frustum 
of  a  rt.  circ.  cone,  §  620  ;  of  a 
cylinder,  defined,  §  597 ;  of  a 
cylinder,  §  601 ;  of  any  parallele- 
piped, §  651 ;  of  any  prism,  §  553 ; 
of  a  solid,  §   633;    of  a  sphere, 


Zone,  §  666. 


UNIVERS 


NIA  LIBRARY 


BERKELEY 

Return  to  desk  from  which  borrowed. 
This  book  is  DUE  on  the  last  date  stamped  below. 


FEB  1 1  13A8 


LD  21-100m-9,'47(A5702sl6)476 


YB  35944 


! 

i 

208 

UNIVERSITY  OF  CAUFORNIA  LIBRARY 

